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Graph of the parabola y=0.5*(x-3)^2 - 4
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 \(\normalsize{a<0}\) 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).

Graph of y=4x^2, y=x^2 and y=x^2/4

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 \(\normalsize{ y=-4}\).

Graph of y=1/4x^2-3, together with its focus and directrix

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 \(\normalsize{[5,0]}\).

Graph of y=0.25*(x-5)^2

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 \(\normalsize F=[0,\frac{1}{4a}]\).

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