It's a matter of perspective

It's a matter of perspective

Excerpt: This resource asks students to find the equation of a parabola given three points. It can be used as a follow-on from 'Name that graph again', and it introduces a wider range of transformations than those previously encountered.

Problem

We would like to find an equation for this parabola in the form $y=f(x)$.

What ways come to mind straight away?

A certain point of view

Click on the Animate button in the applet below or move the slider.

[interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/kgWsjQfY/width/600/height/400/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/false/ld/false/sri/true/asb/false/at/html5] — not part of the text import; see the source page.

Download GeoGebra file

What is happening in the animation as the value of $u$ increases from $0$ to $1$?

As $u$ increases, the points $(0,5)$ and $(2,1)$ are moved in such a way that their $x$-coordinates stay the same but their $y$-coordinates smoothly change to $0$. They are transformed to $(0,0)$ and $(2,0)$.

The red dashed line is drawn passing through the two points described above. This line becomes horizontal (and lies on the $x$-axis) when $u=1$.

The original parabola is transformed so that it always passes through these two points. It appears to remain as a parabola.

The point $C$ marked on the parabola remains fixed throughout.

An answer

Which characteristics of the parabola are being preserved and which are being changed?

The curve does remain as a parabola as $u$ increases, so we could say that its “parabolic” shape stays the same. Its orientation remains upward facing.

However, it is transformed so that when $u=1$ it passes through the points $(0,0)$ and $(2,0)$ rather than the points $(0,5)$ and $(2,1)$. The original parabola does not intersect the $x$-axis but the transformed one does.

An answer

How can we describe the transformation that is happening to the parabola?

Each point on the parabola is being transformed in the same way as each point on the straight line.

The equation of the original straight line can be written as $y=5-2x$. As $u$ increases, we are gradually subtracting from its $y$ values until when $u=1$ the equation of the line has become $y=0$. At this stage, we have therefore subtracted $(5-2x)$ from the $y$ value of each point on the line.

This type of transformation is called a shear; in this case, it is a shear parallel to the $y$-axis. A shear is a particular example of an affine transformation.

Similarly, if we write the equation of the original parabola as

$y=f(x)$

and subtract

$(5-2x)$

when

$u=1$

, we can say that the transformed parabola has equation

$$\begin{equation} y=f(x)-(5-2x). \end{equation}$$

An answer

What can we say about the transformed parabola?

where $K$ is some constant.

We can see that the transformed parabola intersects the

$x$

-axis at

$x=0$

and

$x=2$

, so its equation can be written as

$$\begin{equation} y=Kx(x-2), \end{equation}$$

An answer

Can we use the transformed parabola to find the equation of the original parabola, $y=f(x)$?

We saw above $(eq1)$ that the transformed parabola has equation $y=f(x)-(5-2x)$, and also $(eq2)$ that it can be written as $y=Kx(x-2)$. Therefore $f(x)-(5-2x)=Kx(x-2),$ and we can rearrange this to get $f(x)=5-2x+Kx(x-2).$

and the equation of the original parabola is $y = 2x^2-6x+5.$

We can find the value of

$K$

by using the third given point on the original parabola

$(4,13)$

:

$$13 = 5-2\times4+4K(4-2)$$

and solving this gives

$K = 2$

. Thus

$$\begin{align*} f(x) &= 5-2x+2x(x-2)\\ &= 2x^2-6x+5 \end{align*}$$

An answer

A point $C$ is marked on the diagram. What happens to the point $C$ as $u$ varies and what condition must the point $C$ satisfy?

A final thought