Difference between revisions of "Vertex equations of a quadratic function and it's inverse"
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:<math>y = a (x-v_x)^2 + v_y.</math> | :<math>y = a (x-v_x)^2 + v_y.</math> | ||
− | + | <math>a</math> can be determined by solving | |
− | + | :<math>p_y = a (p_x-v_x)^2 + v_y</math> for <math>a</math> which gives | |
− | + | :<math> a = (p_y - v_y) / (p_x - v_x)^2 .</math> | |
Line 42: | Line 42: | ||
===Inverse quadratic function=== | ===Inverse quadratic function=== | ||
− | Conversely, also the inverse quadratic function can be uniquely defined by its vertex | + | Conversely, also the inverse quadratic function can be uniquely defined by its vertex <math>V</math> and one more point <math>P</math>. |
The function term of the inverse function has the form | The function term of the inverse function has the form | ||
− | + | :<math>y = sqrt((x-v_x)/a) + v_y.</math> | |
− | + | <math>a'' can be determined by solving | |
− | + | :<math>p_y = sqrt((p_x-v_x)/a) + v_y</math> for <math>a</math> which gives | |
− | + | :<math>a = (p_x - v_x) / (p_y - v_y)^2.</math> | |
Revision as of 16:15, 15 January 2021
A parabola can be uniquely defined by its vertex V=(v_x, v_y) and one more point P=(p_x, p_y). The function term of the parabola then has the form
- [math]y = a (x-v_x)^2 + v_y.[/math]
[math]a[/math] can be determined by solving
- [math]p_y = a (p_x-v_x)^2 + v_y[/math] for [math]a[/math] which gives
- [math] a = (p_y - v_y) / (p_x - v_x)^2 .[/math]
JavaScript code
var b = JXG.JSXGraph.initBoard('box1', {boundingbox: [-5, 5, 5, -5], grid:true});
var v = b.create('point', [0,0], {name:'V'}),
p = b.create('point', [3,3], {name:'P'}),
f = b.create('functiongraph', [
function(x) {
var den = p.X()- v.X(),
a = (p.Y() - v.Y()) / (den * den);
return a * (x - v.X()) * (x - v.X()) + v.Y();
}]);
})();
Inverse quadratic function
Conversely, also the inverse quadratic function can be uniquely defined by its vertex [math]V[/math] and one more point [math]P[/math]. The function term of the inverse function has the form
- [math]y = sqrt((x-v_x)/a) + v_y.[/math]
[math]a'' can be determined by solving :\ltmath\gtp_y = sqrt((p_x-v_x)/a) + v_y[/math] for [math]a[/math] which gives
- [math]a = (p_x - v_x) / (p_y - v_y)^2.[/math]
JavaScript code
var b = JXG.JSXGraph.initBoard('box2', {boundingbox: [-5, 5, 5, -5], grid:true});
var v = b.create('point', [0,0], {name:'V'}),
p = b.create('point', [3,3], {name:'P'}),
f = b.create('functiongraph', [
function(x) {
var den = p.Y()- v.Y(),
a = (p.X() - v.X()) / (den * den),
sign = (p.Y() >= 0) ? 1 : -1;
return sign * Math.sqrt((x - v.X()) / a) + v.Y();
}]);