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2.8

## UNSW Sydney

A parabola translated from the origin

# Scaling and translating quadratic functions

We can get more general quadratic functions by scaling and translating the standard equation $\normalsize{y=x^2}$ . Pleasantly, any quadratic function can be obtained in this way.

In this step we will see how scaling and translation affect the graph of a quadratic function.

## Scaling the standard parabola

We have already seen that the shape of $\normalsize{y=ax^2}$ depends on the value of the constant $\normalsize{a}$. We can consider this graph to be a scaling of the particular standard parabola $\normalsize{y=x^2}$; if $\normalsize{a>0}$ then the parabola opens upwards, and if $% $ then the parabola opens downwards.

For large values of $\normalsize{a}$ the parabola is steep and narrow. The closer to $\normalsize{0}$ the value $\normalsize{a}$ gets, the flatter the parabola. Indeed when $\normalsize{a=0}$ the parabola is so flat that we get the straight line conic $y=0$, which could be considered as a special case of the parabola.

Here for comparison are the graphs of $\normalsize{y=4x^2}$ (green), $\normalsize{y=x^2}$ (blue) and $\normalsize{y=\frac{x^2}{4}}$ (red).

## Translating a standard parabola up or down

Now we investigate what happens if we translate a standard parabola $\normalsize{y=ax^2}$ up or down. This is quite easy. Note that however we translate a parabola, its vertex, and focus, and directrix will move in exactly the same way.

Q1 (E): Where is the vertex of the parabola obtained by translating $\normalsize{y=4x^2}$ by $\normalsize{3}$ in the $\normalsize{x}$ direction and by $\normalsize{5}$ in the $\normalsize{y}$ direction?

Geometrically, if we add a constant $\normalsize{k}$ to the equation $\normalsize{y=ax^2}$ then we translate the parabola by $\normalsize{k}$ in the vertical direction. The algebraic operation that shifts a function vertically by $k$ can be thought of as replacing $y$ with $y-k$.

Recall that the parabola $\normalsize y=\frac{1}{4} x^2$ has vertex $\normalsize V=[0,0]$, focus $\normalsize F=[0,1]$, and directrix $\normalsize l: y=-1$. Here is a graph of the parabola $\normalsize{y=\frac{1}{4}x^2-3}$, obtained by translating $\normalsize{y=\frac{1}{4}x^2}$ down by $\normalsize{3}$. So its vertex will be at $\normalsize{[0,-3]}$, its focus will be at $\normalsize{[0,-2]}$, and its directrix will be the line $\normalsize{ y=-4}$.

## Translating a standard parabola left or right

What if we want to shift $\normalsize{y=\frac{1}{4}x^2}$ over $\normalsize{5}$ units to the right? We apply the same strategy as when we were translating lines: replace $x$ with $x-5$ to get

It is easy to check the old vertex $\normalsize{[0,0]}$ has moved to $\normalsize{[5,0]}$.

We could expand the equation to get

## Combining translations horizontal and vertical

If we take the standard parabola $\normalsize{y=ax^2}$ and translate it by $\normalsize{h}$ in the $\normalsize{x}$ direction and by $\normalsize{k}$ in the $\normalsize{y}$ direction, then we obtain

This is the general form of a standard parabola which has been translated by the vector, or directed line segment, $\normalsize{(h,k)}$. This an important expression in the theory of quadratic functions.

Q2 (M): What is the vertex of this parabola?

Q3 (M): What are the focus and directrix of $\normalsize{y=1/4(x-5)^2-2}$?

Q4 (C): What is the focus of the parabola $\normalsize{y=ax^2}$?

## A key fact about parabolas

Can we get every parabola this way – just by taking a standard parabola of the form $\normalsize{y=ax^2}$, and then shifting or translating in the $\normalsize{x}$-direction by a certain amount and then in the $\normalsize{y}$-direction by a certain amount? Yes we can!

This is a pleasant and important fact about parabolas. It shows us that all quadratic functions, even ones with complicated formulas like $\normalsize{y=7x^2-11x+8}$ have essentially similar shapes, and that we can find out where they are situated by unravelling how much $\normalsize{x}$ and $\normalsize{y}$ translations are required to obtain them from standard parabolas.

So there is a fundamental question here: how can we translate a standard parabola to get the general parabola $\normalsize{y=ax^2+bx+c}$? Finding the answer will take us to ancient Persia, to a technique called completing the square, and to a remarkable identity that all students of mathematics ought to have seen.

A1. If we translate a parabola, then its vertex translates correspondingly. Since the vertex of $\normalsize{y=4x^2}$ is the origin $\normalsize{[0,0]}$, the vertex of the translated parabola must be $\normalsize{[3,5]}$.

A2. The unshifted parabola has a vertex at $\normalsize{[0,0]}$, so the new vertex will be at $\normalsize{[h,k]}$.

A3. The unshifted parabola $y=\frac14 x^2$ has focus at $[0,1]$ and directrix $y=-1$. We have essentially just moved the entire picture to the right by $5$ and down by $2$. So visually we can see the focus is now located at $[5,-1]$ and the directrix is now $y=-3$.

A4. The focus of the parabola $\normalsize y=ax^2$ is the point $\normalsize F=[0,\frac{1}{4a}]$.