Embeddable Graphs

Going beyond just polynomials, I’ve created a javascript graphing app that’s easily embeddable.

At the moment, it just does polynomials and points, but polynomials can be used to teach quadratic functions (parabolas) and straight lines to pre-algebra and algebra students. Which I’ve been doing.

Based on my students’ feedback, I’ve made it so that when you change the equation of the line the movement animates. This makes it much easier to see what happens when, for example, you change the slope of a line.

P.S. You can also turn off the interactivity if you just want to show a simple graph. y = x2-1 is shown below:

The Algebra of Straight Lines

A quick graphical calculator for straight lines — based on This. It’s still incomplete, but it’s functional, and a useful complement to the equation of a straight line animation and the parabola graphing.

You can graph lines based on the equation of a line (in either the slope-intercept or point-slope forms), or from two points.

[inline]

Straight Lines


Slope-Intercept Form Two Points Point-Slope Form

y = m x + b

(x1, y1) =
( ,
)

y – y1 = m ( x – x1)
y =
x +


(x2, y2) =
( ,
)
y –
=
( x –
)

x intercept:

Slope:

Equation:
y intercept:

Points:


Your browser does not support the canvas element.

[script type=”text/javascript”]

var width=500;
var height=500;
var xrange=10;
var yrange=10;

mx = width/(2.0*xrange);
bx = width/2.0;
my = -height/(2.0*yrange);
by = height/2.0;

document.getElementById(‘testing’).innerHTML = “Hello2”;

function draw_9359(ctx, polys) {

t_9359=t_9359+dt_9359;
//ctx.fillText (“t=”+t, xp(5), yp(5));
ctx.clearRect(0,0,width,height);

polys[0].drawAxes(ctx);
ctx.lineWidth=2;
polys[0].draw(ctx);

polys[0].write_eqn2(ctx);

//ctx.fillText (polys[0].get_parabola_vertex_form_eqn(), xp(5), yp(-5) );

// SHOW 2 POINTS USED TO CALCULATE LINE
if (show_2pts_ctrl_9359.checked==true) {
polys[0].circle_point(ctx, polys[0].p1.x, polys[0].p1.y, “#000”, 0.15);
polys[0].label_point_on_line(ctx, polys[0].p1.x, polys[0].p1.y);
polys[0].circle_point(ctx, polys[0].p2.x, polys[0].p2.y, “#000”, 0.15);
polys[0].label_point_on_line(ctx, polys[0].p2.x, polys[0].p2.y);

}

// SHOW Y INTERCEPT
if (show_y_intercept_ctrl_9359.checked==true) {
polys[0].draw_y_intercept_line(ctx);
document.getElementById(‘y_intercept_pos_9359’).innerHTML = “( 0 , “+ polys_9359[0].c.toPrecision(2)+ ” )”;

if (polys[0].b > 0.0) {
ctx.textAlign=”right”;
xoff = -0.5;
}
else {
ctx.textAlign=”left”;
xoff = 0.5;
}
ctx.fillText (‘y intercept: y = ‘+polys[0].c.toPrecision(2), xp(xoff), yp(polys_9359[0].c+0.25));

} else { document.getElementById(‘y_intercept_pos_9359’).innerHTML = “”;}

// SHOW SLOPE
if (show_slope_ctrl_9359.checked==true) {
polys[0].draw_slope_line(ctx);
document.getElementById(‘slope_pos_9359’).innerHTML = ” = “+polys[0].b.toPrecision(2);

} else { document.getElementById(‘slope_pos_9359’).innerHTML = “”;}

document.getElementById(‘slope_int_equation_9359’).innerHTML = polys[0].get_eqn2();

//SHOW INTERCEPTS
ctx.textAlign=”center”;
if (show_intercepts_ctrl_9359.checked==true) {
polys[0].x_intercepts(ctx);
if (polys[0].order == 2) {
if (polys[0].x_intcpts.length > 0) {
line = “0 = “;
for (var i=0; i 0.0) { sign=”-“;} else {sign=”+”;}
line = line + “(x “+sign+” “+ Math.abs(polys[0].x_intcpts[i].toPrecision(2))+ “)”;
}
ctx.fillText (line, xp(6), yp(7));
for (var i=0; i 0) {
if (polys[0].b > 0.0) {
ctx.textAlign=”right”;
xoff = -0.5;
}
else {
ctx.textAlign=”left”;
xoff = 0.5;
}
ctx.font = “12pt Calibri”;
ctx.fillStyle = “#00f”;
ctx.fillText (‘x intercept: x = ‘+polys[0].x_intcpts[0].toPrecision(2), xp(polys[0].x_intcpts[0]+xoff), yp(0.5));
document.getElementById(‘intercepts_pos_9359’).innerHTML = “( “+ polys[0].x_intcpts[0].toPrecision(2) + “, 0 )”;
}
else {
ctx.fillText (‘None’, xp(7), yp(7));
}

}

}
else {
document.getElementById(‘intercepts_pos_9359’).innerHTML = ” “;
}

// Finding the slope
outln = “

Finding the Slope

“;
outln += “

The slope of the line between the two points is the rise divided by the run: the change in elevation divided by the horizontal distance between the points.”;

// Write out intercept solution
outln = “

Finding the x-intercept

“;
outln += “

Start with the slope-intercept form of the equation:”;
outln += “

     “+polys[0].get_eqn2(“y”,”x”,”html”)+”“;
outln += “

Know that at the x-intercept, y = 0 , which you substitute into the equation to get:”;
outln += “

     “+polys[0].get_eqn2(“0″,”x”,”html”)+”“;
outln += “

Now move the b coefficient to the other side of the equation:”;
outln += “

     “+(-polys[0].c).toPrecision(2)+” = “+ polys[0].b.toPrecision(2)+”x”+”“;
outln += “

And divide both sides by the slope to solve for x:”;
outln += “

     “+(-polys[0].c).toPrecision(2)+ ” / “+ polys[0].b.toPrecision(2) + ” = “+ polys[0].b.toPrecision(2)+”x”+ ” / “+ polys[0].b.toPrecision(2)+”“;
outln += “

     “+(-polys[0].c/polys[0].b).toPrecision(2)+” = “+”x”+”“;
outln += “

Therefore, the line intercepts the x axis at the point:”;
outln += “

     ( “+(-polys[0].c/polys[0].b).toPrecision(2)+” , 0 )”;
outln += “

When,”;
outln += “

     “+”x = “+ (-polys[0].c/polys[0].b).toPrecision(2) +”“;

document.getElementById(‘x_intercept_9359’).innerHTML = outln;

//Write out the solution by factoring
solution = “

Finding the y-intercept

“;
if (polys[0].order == 2) {
solution = solution + “y = “+polys[0].a.toPrecision(2)+” x2 “+polys[0].bsign+” “+Math.abs(polys[0].b.toPrecision(2))+” x “+ polys[0].csign+” “+Math.abs(polys[0].c.toPrecision(2))+”

“;

solution = solution + ‘Factoring:      ‘;

if (polys[0].x_intcpts.length > 0) {
solution = solution + ‘0 = ‘;
for (var i=0; i 0.0) { sign=”-“;} else {sign=”+”;}
solution = solution + “(x “+sign+” “+ Math.abs(polys[0].x_intcpts[i].toPrecision(2))+ “)”;
}
solution = solution + ‘

‘;
solution = solution + ‘Set each factor equal to zero:
     ‘;
for (var i=0; i 0.0) { sign=”-“;} else {sign=”+”;}
solution = solution + “x “+sign+” “+ Math.abs(polys[0].x_intcpts[i].toPrecision(2))+ ” = 0           “;
}
solution = solution + ‘

and solve for x:
     ‘;
for (var i=0; i‘;
}
document.getElementById(‘equation_9359’).innerHTML = solution;
}

else if (polys[0].order == 1) {
solution = solution + ‘

At the y-axis x = 0.
To find the y intercept, take the equation of the line, set x = 0, and solve for x.’;
solution = solution + ‘

     ‘;
solution = solution + “y = “+” “+Math.abs(polys[0].b.toPrecision(2))+” x “+ polys[0].csign+” “+Math.abs(polys[0].c.toPrecision(2))+”

“;
solution = solution + ‘     ‘;
solution = solution + ” 0 = “+” “+Math.abs(polys[0].b.toPrecision(2))+” x “+ polys[0].csign+” “+Math.abs(polys[0].c.toPrecision(2))+”

“;
solution = solution + ‘     ‘;
solution = solution + (-1.0*polys[0].c).toPrecision(2) +” = “+” “+Math.abs(polys[0].b.toPrecision(2))+” x “+”

“;
solution = solution + ‘     ‘;
solution = solution + (-1.0*polys[0].c).toPrecision(2)+”/”+polys[0].b.toPrecision(2)+” = “+” x “+”

“;
solution = solution + ‘     ‘;
solution = solution + “x = “+ (-1.0*polys[0].c/polys[0].b).toPrecision(4)+”

“;

document.getElementById(‘equation_9359’).innerHTML = solution;

}

}

function update_form_9359 () {
b_coeff_9359.value = polys_9359[0].b+””;
c_coeff_9359.value = polys_9359[0].c+””;

y1_coeff_9359.value = polys_9359[0].y1+””;
b1_coeff_9359.value = polys_9359[0].b+””;
x1_coeff_9359.value = polys_9359[0].x1+””;

p1_x_9359.value = polys_9359[0].p1.x+””;
p1_y_9359.value = polys_9359[0].p1.y+””;
p2_x_9359.value = polys_9359[0].p2.x+””;
p2_y_9359.value = polys_9359[0].p2.y+””;
}

//init_mouse();

var c_9359=document.getElementById(“myCanvas_9359”);
var ctx_9359=c_9359.getContext(“2d”);

var change = 0.0001;

function create_lines_9359 () {
//draw line
//document.write(“hello world! “);
var polys = [];
polys.push(addPoly(0,2, 3));
document.getElementById(‘testing’).innerHTML = “Hello”;
polys[0].set_2_points_line();
document.getElementById(‘testing’).innerHTML = “Hello4″;

// polys[1].color = ‘#8C8’;

return polys;
}

var polys_9359 = create_lines_9359();

var x1=xp(-10);
var y1=yp(1);
var x2=xp(10);
var y2=yp(1);
var dc_9359=0.05;

var t_9359 = 0;
var dt_9359 = 100;
//end_ct = 0;
var st_pt_x_9359 = 2;
var st_pt_y_9359 = 1;
var show_y_intercept_9359 = 1; //1 to show y intercept on startup
var show_intercepts_9359 = 0; // 1 to show the intercepts
var show_slope_9359 = 0; // 1 to show the slope
var show_2pts_9359 = 0; // 1 to show the two points from which line can be calculated

var move_dir_9359 = 1.0; // 1 for up

//document.getElementById(‘comment_spot’).innerHTML = polys_9359[0].a+” “+polys_9359[0].b+” “+polys_9359[0].c+” : “+polys_9359[0].h+” “+polys_9359[0].k+” “;

document.getElementById(‘testing’).innerHTML = “Hello3”;

//slope-intercept form
var b_coeff_9359 = document.getElementById(“b_coeff_9359”);
var c_coeff_9359 = document.getElementById(“c_coeff_9359”);

//point-slope form
var y1_coeff_9359 = document.getElementById(“y1_coeff_9359”);
var b1_coeff_9359 = document.getElementById(“b1_coeff_9359”);
var x1_coeff_9359 = document.getElementById(“x1_coeff_9359”);

//equation from two points
var p1_x_9359 = document.getElementById(“x1_9359”);
var p1_y_9359 = document.getElementById(“y1_9359”);
var p2_x_9359 = document.getElementById(“x2_9359”);
var p2_y_9359 = document.getElementById(“y2_9359”);

//options
var show_y_intercept_ctrl_9359 = document.getElementById(“show_y_intercept_9359”);
if (show_y_intercept_9359 == 0) {show_y_intercept_ctrl_9359.checked=false;
} else {show_y_intercept_ctrl_9359.checked=true;}

var show_intercepts_ctrl_9359 = document.getElementById(“show_intercepts_9359”);
if (show_intercepts_9359 == 0) {show_intercepts_ctrl_9359.checked=false;
} else {show_intercepts_ctrl_9359.checked=true;}

var show_slope_ctrl_9359 = document.getElementById(“show_slope_9359”);
if (show_slope_9359 == 0) {show_slope_ctrl_9359.checked=false;
} else {show_slope_ctrl_9359.checked=true;}

var show_2pts_ctrl_9359 = document.getElementById(“show_pts_9359”);
if (show_2pts_9359 == 0) {show_2pts_ctrl_9359.checked=false;
} else {show_2pts_ctrl_9359.checked=true;}

update_form_9359();

//document.write(“test= “+c_coeff_9359.value+” “+polys_9359[0].c);
setInterval(“draw_9359(ctx_9359, polys_9359)”, dt_9359);

b_coeff_9359.onchange = function() {
polys_9359[0].set_b(parseFloat(this.value));
polys_9359[0].update_point_slope_form_line();
polys_9359[0].set_2_points_line();
update_form_9359();
}
c_coeff_9359.onchange = function() {
//polys_9359[0].c = parseFloat(this.value);
polys_9359[0].set_c(parseFloat(this.value));
polys_9359[0].update_point_slope_form_line();
polys_9359[0].set_2_points_line();
update_form_9359();
}

y1_coeff_9359.onchange = function() {
//polys_9359[0].a = parseFloat(this.value);
polys_9359[0].y1 = parseFloat(this.value);
polys_9359[0].update_slope_intercept_form_line();
polys_9359[0].set_2_points_line();
update_form_9359();
}

b1_coeff_9359.onchange = function() {
polys_9359[0].set_b(parseFloat(this.value));
polys_9359[0].update_slope_intercept_form_line();
polys_9359[0].set_2_points_line();
update_form_9359();
}

x1_coeff_9359.onchange = function() {
polys_9359[0].x1 = parseFloat(this.value);
polys_9359[0].update_slope_intercept_form_line();
polys_9359[0].set_2_points_line();
polys_9359[0].set_order()
update_form_9359();
}

p1_x_9359.onchange = function() {
polys_9359[0].p1.x = parseFloat(this.value);
polys_9359[0].update_line_from_2pts();
polys_9359[0].set_order()
update_form_9359();
}

p2_x_9359.onchange = function() {
polys_9359[0].p2.x = parseFloat(this.value);
polys_9359[0].update_line_from_2pts();
polys_9359[0].set_order()
update_form_9359();
}

p1_y_9359.onchange = function() {
polys_9359[0].p1.y = parseFloat(this.value);
polys_9359[0].update_line_from_2pts();
polys_9359[0].set_order()
update_form_9359();
}

p2_y_9359.onchange = function() {
polys_9359[0].p2.y = parseFloat(this.value);
polys_9359[0].update_line_from_2pts();
polys_9359[0].set_order()
update_form_9359();
}

//draw_9359();
//document.write(“x”+x2+”x”);
//ctx_9359.fillText (“n=”, xp(5), yp(5));

[/script]

[/inline]

Everything You (N)ever Wanted to Know About Parabolas

So that my students could more easily check their answers graphically, I put together a page with a more complete analysis of parabolas (click this link for more details).

[inline]

Analyzing Parabolas


Standard Form Vertex Form

y = a x2 + b x + c

y = a (x – h)2 + k
y =
x2 +
x +

y =
( x –
) 2 +

Intercepts:

Vertex:
Focus:

Directrix:

Axis:


Your browser does not support the canvas element.

Solution by Factoring:

y = x2 x

[script type=”text/javascript”]
var width=500;
var height=500;
var xrange=10;
var yrange=10;

mx = width/(2.0*xrange);
bx = width/2.0;
my = -height/(2.0*yrange);
by = height/2.0;

function draw_9239(ctx, polys) {
t_9239=t_9239+dt_9239;
//ctx.fillText (“t=”+t, xp(5), yp(5));
ctx.clearRect(0,0,width,height);

polys[0].drawAxes(ctx);
ctx.lineWidth=2;
polys[0].draw(ctx);

polys[0].write_eqn2(ctx);

//polys[0].y_intercepts(ctx);
//write intercepts on graph

// SHOW VERTEX
if (show_vertex_ctrl_9239.checked==true) {
polys[0].draw_vertex(ctx);
document.getElementById(‘vertex_pos_9239’).innerHTML = “(“+polys_9239[0].vertex.x.toPrecision(2)+ ” , “+polys_9239[0].vertex.y.toPrecision(2)+ ” )”;
} else { document.getElementById(‘vertex_pos_9239’).innerHTML = “”;}

// SHOW FOCUS
if (show_focus_ctrl_9239.checked==true) {
polys[0].draw_focus(ctx);
document.getElementById(‘focus_pos_9239’).innerHTML = “(“+polys_9239[0].focus.x.toPrecision(2)+ ” , “+polys_9239[0].focus.y.toPrecision(2)+ ” )”;
} else { document.getElementById(‘focus_pos_9239’).innerHTML = “”;}

// SHOW DIRECTRIX
if (show_directrix_ctrl_9239.checked==true) {
polys[0].draw_directrix(ctx);
document.getElementById(‘directrix_pos_9239’).innerHTML = polys[0].directrix.get_eqn2(“y”,”x”,”html”);

polys[0].directrix.write_eqn2(ctx, polys[0].directrix.get_eqn2(“directrix: y”));
} else { document.getElementById(‘directrix_pos_9239’).innerHTML = “”;}

// SHOW AXIS
if (show_axis_ctrl_9239.checked==true) {
polys[0].draw_parabola_axis(ctx);
document.getElementById(‘axis_pos_9239’).innerHTML = “x = “+polys[0].vertex.x.toPrecision(2);
}

//SHOW INTERCEPTS
ctx.textAlign=”center”;
if (show_intercepts_ctrl_9239.checked==true) {
polys[0].x_intercepts(ctx);
ctx.fillText (‘x intercepts: (when y=0)’, xp(6), yp(8));
//ctx.fillText (‘intercepts=’+polys[0].x_intcpts.length, xp(5), yp(-5));
if (polys[0].order == 2) {
if (polys[0].x_intcpts.length > 0) {
line = “0 = “;
for (var i=0; i 0.0) { sign=”-“;} else {sign=”+”;}
line = line + “(x “+sign+” “+ Math.abs(polys[0].x_intcpts[i].toPrecision(2))+ “)”;
}
ctx.fillText (line, xp(6), yp(7));
for (var i=0; i 0) {
for (var i=0; i2 “+polys[0].bsign+” “+Math.abs(polys[0].b.toPrecision(2))+” x “+ polys[0].csign+” “+Math.abs(polys[0].c.toPrecision(2))+”

“;

solution = solution + ‘Factoring:      ‘;

if (polys[0].x_intcpts.length > 0) {
solution = solution + ‘0 = ‘;
for (var i=0; i 0.0) { sign=”-“;} else {sign=”+”;}
solution = solution + “(x “+sign+” “+ Math.abs(polys[0].x_intcpts[i].toPrecision(2))+ “)”;
}
solution = solution + ‘

‘;
solution = solution + ‘Set each factor equal to zero:
     ‘;
for (var i=0; i 0.0) { sign=”-“;} else {sign=”+”;}
solution = solution + “x “+sign+” “+ Math.abs(polys[0].x_intcpts[i].toPrecision(2))+ ” = 0           “;
}
solution = solution + ‘

and solve for x:
     ‘;
for (var i=0; i‘;
}
document.getElementById(‘equation_9239’).innerHTML = solution;
}

else if (polys[0].order == 1) {
solution = solution + ‘
     ‘;
solution = solution + “y = “+” “+Math.abs(polys[0].b.toPrecision(2))+” x “+ polys[0].csign+” “+Math.abs(polys[0].c.toPrecision(2))+”

“;
solution = solution + ‘

Set y=0 and solve for x:
     ‘;
solution = solution + ” 0 = “+” “+Math.abs(polys[0].b.toPrecision(2))+” x “+ polys[0].csign+” “+Math.abs(polys[0].c.toPrecision(2))+”

“;
solution = solution + ‘     ‘;
solution = solution + (-1.0*polys[0].c).toPrecision(2) +” = “+” “+Math.abs(polys[0].b.toPrecision(2))+” x “+”

“;
solution = solution + ‘     ‘;
solution = solution + (-1.0*polys[0].c).toPrecision(2)+”/”+polys[0].b.toPrecision(2)+” = “+” x “+”

“;
solution = solution + ‘     ‘;
solution = solution + “x = “+ (-1.0*polys[0].c/polys[0].b).toPrecision(4)+”

“;

document.getElementById(‘equation_9239’).innerHTML = solution;
}

}

function update_form_9239 () {
a_coeff_9239.value = polys_9239[0].a+””;
b_coeff_9239.value = polys_9239[0].b+””;
c_coeff_9239.value = polys_9239[0].c+””;

av_coeff_9239.value = polys_9239[0].a+””;
hv_coeff_9239.value = polys_9239[0].h+””;
kv_coeff_9239.value = polys_9239[0].k+””;

}

//init_mouse();

var c_9239=document.getElementById(“myCanvas_9239”);
var ctx_9239=c_9239.getContext(“2d”);

var change = 0.0001;

function create_lines_9239 () {
//draw line
//document.write(“hello world! “);
var polys = [];
polys.push(addPoly(1,6, 5));

// polys.push(addPoly(0.25, 1, 0));
// polys[1].color = ‘#8C8’;

return polys;
}

var polys_9239 = create_lines_9239();

var x1=xp(-10);
var y1=yp(1);
var x2=xp(10);
var y2=yp(1);
var dc_9239=0.05;

var t_9239 = 0;
var dt_9239 = 100;
//end_ct = 0;
var st_pt_x_9239 = 2;
var st_pt_y_9239 = 1;
var show_vertex_9239 = 1; //1 to show vertex on startup
var show_focus_9239 = 1; // 1 to show the focus
var show_intercepts_9239 = 1; // 1 to show the intercepts
var show_directrix_9239 = 1; // 1 to show the directrix
var show_axis_9239 = 1; //1 to show the axis of the parabola

var move_dir_9239 = 1.0; // 1 for up

//document.getElementById(‘comment_spot’).innerHTML = polys_9239[0].a+” “+polys_9239[0].b+” “+polys_9239[0].c+” : “+polys_9239[0].h+” “+polys_9239[0].k+” “;

//standard form
var a_coeff_9239 = document.getElementById(“a_coeff_9239”);
var b_coeff_9239 = document.getElementById(“b_coeff_9239”);
var c_coeff_9239 = document.getElementById(“c_coeff_9239”);

//vertex form
var av_coeff_9239 = document.getElementById(“av_coeff_9239”);
var hv_coeff_9239 = document.getElementById(“hv_coeff_9239”);
var kv_coeff_9239 = document.getElementById(“kv_coeff_9239”);

//options
var show_vertex_ctrl_9239 = document.getElementById(“show_vertex_9239”);
if (show_vertex_9239 == 0) {show_vertex_ctrl_9239.checked=false;
} else {show_vertex_ctrl_9239.checked=true;}

var show_focus_ctrl_9239 = document.getElementById(“show_focus_9239”);
if (show_focus_9239 == 0) {show_focus_ctrl_9239.checked=false;
} else {show_focus_ctrl_9239.checked=true;}

var show_intercepts_ctrl_9239 = document.getElementById(“show_intercepts_9239”);
if (show_intercepts_9239 == 0) {show_intercepts_ctrl_9239.checked=false;
} else {show_intercepts_ctrl_9239.checked=true;}

var show_directrix_ctrl_9239 = document.getElementById(“show_directrix_9239”);
if (show_directrix_9239 == 0) {show_directrix_ctrl_9239.checked=false;
} else {show_directrix_ctrl_9239.checked=true;}

var show_axis_ctrl_9239 = document.getElementById(“show_axis_9239”);
if (show_axis_9239 == 0) {show_axis_ctrl_9239.checked=false;
} else {show_axis_ctrl_9239.checked=true;}

update_form_9239();

//document.write(“test= “+c_coeff_9239.value+” “+polys_9239[0].c);
setInterval(“draw_9239(ctx_9239, polys_9239)”, dt_9239);

a_coeff_9239.onchange = function() {
//polys_9239[0].a = parseFloat(this.value);
polys_9239[0].set_a(parseFloat(this.value));
polys_9239[0].update_vertex_form_parabola();
update_form_9239();
}
b_coeff_9239.onchange = function() {
//polys_9239[0].b = parseFloat(this.value);
polys_9239[0].set_b(parseFloat(this.value));
polys_9239[0].update_vertex_form_parabola();
update_form_9239();
}
c_coeff_9239.onchange = function() {
//polys_9239[0].c = parseFloat(this.value);
polys_9239[0].set_c(parseFloat(this.value));
polys_9239[0].update_vertex_form_parabola();
update_form_9239();
}

av_coeff_9239.onchange = function() {
//polys_9239[0].a = parseFloat(this.value);
polys_9239[0].set_a(parseFloat(this.value));
polys_9239[0].update_standard_form_parabola();
update_form_9239();
}

hv_coeff_9239.onchange = function() {
polys_9239[0].h = parseFloat(this.value);
polys_9239[0].update_standard_form_parabola();
polys_9239[0].set_order()
update_form_9239();
}

kv_coeff_9239.onchange = function() {
polys_9239[0].k = parseFloat(this.value);
polys_9239[0].update_standard_form_parabola();
polys_9239[0].set_order()
update_form_9239();
}

//draw_9239();
//document.write(“x”+x2+”x”);
//ctx_9239.fillText (“n=”, xp(5), yp(5));

[/script]

[/inline]

Converting the forms

The key relationships are the ones to convert from the standard form of the parabolic equation:

         y = a x^2 + b x + c (1)

to the vertex form:

         y = a (x - h)^2 + k (2)

If you multiply out the vertex equation form you get:

         y = a x2 – 2ah x + ah2 + k (3)

When you compare this equation to the standard form of the equation (Equation 1), if you look at the coefficients and the constants, you can see that:

To convert from the vertex to the standard form use:

          a = a (4)
          b = -2ah (5)
          c = ah^2 + k (6)

Going the other way,

To convert from the standard to the vertex form of parabolic equations use:

          a = a

(7)
          h = \frac{-b}{2a}

(8)
          k = c - ah^2

(9)

Although it is sometimes convenient to let k not depend on coefficients from its own equation:

          k = c - \frac{b^2}{4a} (10)

The Vertex and the Axis

The nice thing about the vertex form of the equation of the parabola is that if you want the find the coordinates of the vertex of the parabola, they’re right there in the equation.

Specifically, the vertex is located at the point:

          (x_v, y_v) = (h, k) (11)

The axis of the parabola is the vertical line going through the vertex, so:

The equation for the axis of a parabola is:

          x = h (12)

Focus and Directrix

Finally, it’s important to note that the distance (d) from the vertex of the parabola to its focus is given by:

          d = \frac{1}{4a} (13)

Which you can just add d on to the coordinates of the vertex (Equation 11) to get the location of the focus.

          (x_f, y_f) = (x_v, y_v + d)  (14)

The directrix is just the opposite, vertical distance away, so the equation for the directrix is the equation of the horizontal line at:

          y = y_v + d  (15)

References

There are already some excellent parabola references out there including:

Euler’s Method for Approximating a Solution to a Differential Equation

Based on Euler’s Method, this interactive graph illustrates a numerical method for solving differential equations. This approach is at the core of many sophisticated computer models of physical phenomena (like climate and weather).

[inline]

  • Starting point: (x,y) = (   ,   )
  • Step size:
  • Direction:


Your browser does not support the canvas element.

  • Slope equation: dy/dx =   x +   
  • Show analytical solution:

If you know the equation for the slope of a curve (the red line for example),

and a point that the curve passes through, such as , you can integrate to find the equation of the curve:

[script type=”text/javascript”]

var width=500;
var height=500;
var xrange=10;
var yrange=10;

mx = width/(2.0*xrange);
bx = width/2.0;
my = -height/(2.0*yrange);
by = height/2.0;

function upause () {
ct = 1;
}

function draw_9110(ctx, polys) {
t_9110=t_9110+dt_9110;
//ctx.fillText (“t=”+t, xp(5), yp(5));
ctx.clearRect(0,0,width,height);

polys[0].drawAxes(ctx);
polys[0].slopeField(ctx, 1.0, 1.0, 0.5);
ctx.lineWidth=2;
polys[0].draw(ctx);
polys[0].write_eqn(ctx, “slope = dy/dx”);

eu = eu_9110;
//override nsteps values
//max_nsteps_9110 = eu.nsteps;
//eu.nsteps = nsteps_9110;

if (nsteps_9110 <= max_nsteps_9110) { nsteps_9110 = nsteps_9110 + 1;} else { nsteps_9110 = 0; } eu.nsteps = nsteps_9110; if (show_parabola_9110 == 1) { polys[0].Integrate(ctx, eu.x1, eu.y1); polys[0].integral.color = '#3c3'; polys[0].integral.draw(ctx); } pos = polys[0].EulerApprox(ctx, eu); //document.getElementById('comment_9110').innerHTML = eu.nsteps+ " " + eu.x1 + " " + eu.y1+ " " + eu.dx+ " " + eu.dir+ " " + max_nsteps_9110; //starting point label ctx.textAlign="right"; ctx.textBaseline="alphabetic"; ctx.font = "14pt Calibri"; ctx.fillStyle = "#00f"; ctx.fillText ('start = ('+eu.x1+','+eu.y1+")", xp(eu.x1-0.25), yp(eu.y1-0.5)); //pos = polys[0].EulerApprox(ctx, st_pt_x_9110, st_pt_y_9110, 0.5,20,-1); //pos = polys[0].EulerApprox(ctx, st_pt_x_9110, st_pt_y_9110, 1, 6,1); //draw starting point ctx.strokeStyle="#00f"; ctx.lineWidth=2; ctx.beginPath(); ctx.arc(xp(eu.x1),yp(eu.y1),dxp(0.1),0,Math.PI*2); ctx.closePath(); ctx.stroke(); //ctx.fillText (' c='+move_dir_9110+' '+polys[1].c, xp(5), yp(5)); } //init_mouse(); var c_9110=document.getElementById("myCanvas_9110"); var ctx_9110=c_9110.getContext("2d"); var change = 0.0001; function create_lines_9110 () { //draw line var polys = []; polys.push(addPoly(0, 0.5, -1)); return polys; } function set_max_nsteps_9110 (eu) { max_nsteps_9110 = 1.5*(xrange - eu.x1)/eu.dx; } var polys_9110 = create_lines_9110(); var x1=xp(-10); var y1=yp(1); var x2=xp(10); var y2=yp(1); var dc_9110=0.05; var t_9110 = 0; var dt_9110 = 500; var st_dx_9110 = 1.0; var st_nsteps_9110 = 0; var nsteps_9110 = 0; var max_nsteps_9110 = 20; var dir_9110 = 1; var show_parabola_9110 = 1; var st_pt_x_9110 = -2; var st_pt_y_9110 = -3; var move_dir_9110 = 1.0; // 1 for up //Form interaction function update_form_9110 (x, y, dx, nsteps, dir, b, c, show_para) { x_9110.value = st_pt_x_9110+""; y_9110.value = st_pt_y_9110+""; dx_9110.value = st_dx_9110.toPrecision(2); //c_max_nsteps_9110.value = max_nsteps_9110+""; //document.getElementById('comment_9110').innerHTML = c_dir_9110; if (dir == 1) { c_dir_9110.selectedIndex = 0; } else {c_dir_9110.selectedIndex = 1;} b_9110.value = b+""; c_9110.value = c+""; if (show_para == 1) { ctrl_show.checked = 1; } else {ctrl_show.checked = 0;} } function get_euler_form_9110 () { pos = {x : parseFloat(x_9110.value), y : parseFloat(y_9110.value)} if (c_dir_9110.selectedIndex == 0) { dir = 1;} else {dir = -1;} eu = addEuler(pos.x, pos.y, parseFloat(dx_9110.value), max_nsteps_9110, dir); return eu } function update_solution_9110 (ctx, eu) { //update equations and solution document.getElementById('slope_eqn_9110').innerHTML = polys_9110[0].get_eqn("dy/dx"); //document.getElementById('eqn_9110').innerHTML = "hey"+eu.y1; polys_9110[0].Integrate(ctx, eu.x1, eu.y1); document.getElementById('eqn_9110').innerHTML = polys_9110[0].integral.get_eqn(); document.getElementById('init_pt_9110').innerHTML = "("+eu.x1+","+eu.y1+")"; } function add_euler_line() { eu_9110 = get_euler_form_9110 (); max_nsteps_9110 = eu_9110.nsteps; nsteps_9110 = 0; } var x_9110 = document.getElementById("x_9110"); var y_9110 = document.getElementById("y_9110"); var dx_9110 = document.getElementById("dx_9110"); //var c_max_nsteps_9110 = document.getElementById("nsteps_9110"); var c_dir_9110 = document.getElementById("dir_9110"); var b_9110 = document.getElementById("b_9110"); var c_9110 = document.getElementById("c_9110"); var ctrl_show = document.getElementById("chk_show_para_9110"); update_form_9110(x_9110, y_9110, dx_9110, nsteps_9110, dir_9110, polys_9110[0].b, polys_9110[0].c, show_parabola_9110); var eu_9110 = get_euler_form_9110 (); set_max_nsteps_9110(eu_9110); update_solution_9110(ctx_9110, eu_9110); setInterval("draw_9110(ctx_9110, polys_9110)", dt_9110); x_9110.onchange = function() { eu_9110 = get_euler_form_9110 (); max_nsteps_9110 = eu_9110.nsteps; nsteps_9110 = 0; } y_9110.onchange = function() { eu_9110 = get_euler_form_9110 (); max_nsteps_9110 = eu_9110.nsteps; nsteps_9110 = 0; } c_dir_9110.onchange = function() { eu_9110 = get_euler_form_9110 (); max_nsteps_9110 = eu_9110.nsteps; nsteps_9110 = 0; } dx_9110.onchange = function() { eu_9110 = get_euler_form_9110 (); max_nsteps_9110 = eu_9110.nsteps; //max_nsteps_9110 = 1.4 * (10 - eu_9110.x1)/eu_9110.dx; set_max_nsteps_9110(eu_9110); nsteps_9110 = 0; } b_9110.onchange = function() { polys_9110[0].set_b(parseFloat(this.value)); eu_9110 = get_euler_form_9110 (); max_nsteps_9110 = eu_9110.nsteps; nsteps_9110 = 0; } c_9110.onchange = function() { polys_9110[0].set_c(parseFloat(this.value)); eu_9110 = get_euler_form_9110 (); max_nsteps_9110 = eu_9110.nsteps; nsteps_9110 = 0; } ctrl_show.onchange = function() { if (this.checked == 1) { show_parabola_9110 = 1;} else {show_parabola_9110 = 0} } [/script] [/inline]

If you don’t have a starting point (initial condition), you can draw a slope field to see what the general pattern of all the possible solutions.

Even with a starting point, however, there are just times when you can’t integrate the slope equation — it’s either too difficult or even impossible.

Then, what you can do is come up with an approximation of what the curve looks like by projecting along the slope from the starting point.

The program above demonstrates how it’s done. This approach is called Euler’s Method, and is gives a numerical approximation rather than finding the exact, analytical solution using calculus (integration).

So why use an approximation when you can find the exact solution? Because, there are quite a number of problems that are impossible or extremely difficult to solve analytically, things like: the diffusion of pollution in a lake; how changing temperature in the atmosphere gives you weather and climate; the flow of groundwater in aquifers; stresses on structural members of buildings; and the list goes on and on.

As with most types of numerical approximations, you get better results if you can reduce the step size between projections of the slope. Try changing the numbers and see.

A more detailed version, with solutions, is here: Euler’s Method.

A good reference: Euler’s Method by Paul Dawkins.

Slope Fields

[inline]

Your browser does not support the canvas element.

[script type=”text/javascript”]
var width=500;
var height=500;
var xrange=10;
var yrange=10;

mx = width/(2.0*xrange);
bx = width/2.0;
my = -height/(2.0*yrange);
by = height/2.0;

function draw9083(ctx, polys) {
t=t+dt;
//ctx.fillText (“t=”+t, xp(5), yp(5));
ctx.clearRect(0,0,width,height);

polys[0].drawAxes(ctx);
polys[0].slopeField(ctx, 1.0, 1.0, 0.5);
ctx.lineWidth=2;
polys[0].draw(ctx);
polys[0].write_eqn(ctx, “dy/dx “);

}

//init_mouse();

var c9083=document.getElementById(“myCanvas9083”);
var ctx9083=c9083.getContext(“2d”);

var change = 0.0001;

function create_lines9083 () {
//draw line
//document.write(“hello world! “);
var polys = [];
polys.push(addPoly(0, 0.5, 1));

return polys;
}

var polys9083 = create_lines9083();

var x1=xp(-10);
var y1=yp(1);
var x2=xp(10);
var y2=yp(1);
var dy=0.01;

var t = 0;
var dt = 50;
end_ct = 0;

var move_dir = 1; // 1 for up

//draw9083();
//document.write(“x”+x2+”x”);
//ctx9083.fillText (“n=”, xp(5), yp(5));
setInterval(“draw9083(ctx9083, polys9083)”, dt);

[/script]
[/inline]

Say you have the equation that gives you the slope of a curve (let  \frac{dy}{dx} ) be the slope):
! \frac{dy}{dx} = \frac{1}{2} x + 1

When you use integration to solve the equation, there are quite the number of possible solutions (infinite actually), because when you integrate:

! y = \int (\frac{1}{2} x + 1 ) dx

you get:
! y = \frac{1}{4} x^2 + x + c

where c is a constant. Unfortunately, you don’t know what c is without more information; it could be anything.

However, even without integrating, we can get a feel for what the curve will look like by plotting what the slope will look like at a bunch of different points in space. This comes in really handy when you end up with a equation for slope that is really hard — or even impossible — to solve.

The graph below show a curve of possible solutions to the slope equation. You should be able to see, as the graph slowly moves up and down, how the slope of the graph corresponds to the slope field.

[inline]

Your browser does not support the canvas element.

[script type=”text/javascript”]
var width=500;
var height=500;
var xrange=10;
var yrange=10;

mx = width/(2.0*xrange);
bx = width/2.0;
my = -height/(2.0*yrange);
by = height/2.0;

function draw9083b(ctx, polys) {
t9083b=t9083b+dt9083b;
//ctx.fillText (“t=”+t, xp(5), yp(5));
ctx.clearRect(0,0,width,height);

polys[0].drawAxes(ctx);
polys[0].slopeField(ctx, 1.0, 1.0, 0.5);
ctx.lineWidth=2;
polys[0].draw(ctx);
polys[0].write_eqn(ctx, “dy/dx “);

if (polys[1].c > yrange) {
move_dir9083b = -1.0;
polys[1].c = yrange;
}
else if (polys[1].c < -yrange) { move_dir9083b = 1.0; polys[1].c = -yrange; } polys[1].c = polys[1].c + dc9083b*move_dir9083b; polys[1].draw(ctx); polys[1].write_eqn(ctx); //ctx.fillText (' c='+move_dir9083b+' '+polys[1].c, xp(5), yp(5)); } //init_mouse(); var c9083b=document.getElementById("myCanvas9083b"); var ctx9083b=c9083b.getContext("2d"); var change = 0.0001; function create_lines9083b () { //draw line //document.write("hello world! "); var polys = []; polys.push(addPoly(0, 0.5, 1)); polys.push(addPoly(0.25, 1, 0)); polys[1].color = '#8C8'; return polys; } var polys9083b = create_lines9083b(); var x1=xp(-10); var y1=yp(1); var x2=xp(10); var y2=yp(1); var dc9083b=0.05; var t9083b = 0; var dt9083b = 100; //end_ct = 0; var move_dir9083b = 1.0; // 1 for up //draw9083b(); //document.write("x"+x2+"x"); //ctx9083b.fillText ("n=", xp(5), yp(5)); setInterval("draw9083b(ctx9083b, polys9083b)", dt9083b); [/script] [/inline]

Numerical Integration

This is an attempt to illustrate numerical integration by animating an HTML5’s canvas.

We’re trying to find the area between x = 1 and x = 5, beneath the parabola:

 y = -\frac{1}{4} x^2 + x + 4

By integrating, the area under the curve can be calculated as being 17 ⅔ (see below for the analytical solution). For numerical integration, however, the area under the curve is filled with trapezoids and the total area is calculated from the sum of all the areas. As you increase the number of trapezoids, the approximation becomes more accurate. The reduction in the error can be seen on the graph: with 1 trapezoid there is a large gap between the shaded area and the curve; more trapezoids fill in the gap better and better.

The table below show how the error (defined as the difference between the calculation using trapezoids and the analytic solution) gets smaller with increasing numbers of trapezoids (n).

Number of trapezoids Area (units2) Error (difference from 17.66)
1 15.00 2.66
2 17.00 0.66
3 17.37 0.29
4 17.50 0.16
5 17.56 0.10
6 17.59 0.07
7 17.61 0.05
8 17.63 0.03
9 17.63 0.03
10 17.64 0.02

Analytic solution

The area under the curve, between x = 1 and x = 5 can be figured out analytically by integrating between these limits.

 Area = \int_{_1}^{^5} \left(-\frac{x^2}{4}  + x + 4 \right) \,dx

 Area = \left[-\frac{x^3}{12}  + \frac{x^2}{2} + 4x \right]_1^5

 Area = \left[-\frac{(5)^3}{12}  + \frac{(5)^2}{2} + 4(5) \right] - \left[-\frac{(1)^3}{12}  + \frac{(1)^2}{2} + 4(1) \right]

 Area = \left[-\frac{125}{12}  + \frac{25}{2} + 16 \right] - \left[-\frac{1}{12}  + \frac{1}{2} + 4 \right]

 Area = \left[ 22 \frac{1}{12} \right]  - \left[4 \frac{5}{12} \right]

 Area = 17 \frac{2}{3}  = 17.6\bar{6}

(See also WolframAlpha’s solution).

Update

Demonstration of this numerical integration using four trapezoids in embeddable graphs.