Parabolas Using Transformations
Theory
Every parabola is a transformed copy of \(y = x^2\). In turning-point form \(y = a(x - h)^2 + k\), the turning point is \((h,\ k)\), and \(a\) dilates the curve and, if negative, reflects it. This Year 10 maths page shows how to describe the transformations, sketch a parabola with its intercepts, and work backwards from a graph or a description to the equation.
A translation slides a graph without changing its shape. Replacing \(x\) by \(x - h\) moves the graph \(h\) units right; adding \(k\) to the whole right-hand side moves it \(k\) units up.
A vertical dilation by factor \(a\) multiplies every \(y\)-value by \(a\): \(y = ax^2\) is narrower than \(y = x^2\) if \(a \gt 1\) and wider if \(0 \lt a \lt 1\). A reflection in the \(x\)-axis changes the sign of every \(y\)-value: \(y = -x^2\).
The turning-point form \(y = a(x - h)^2 + k\) shows all of these at once. The turning point is \((h,\ k)\); it is a minimum if \(a \gt 0\) and a maximum if \(a \lt 0\).
Moving a dilated parabola: \(y = ax^2\) moved \(h\) units right and \(k\) units up is \(y = a(x - h)^2 + k\). The translation does not change \(a\), so the shape stays the same.
Completing the square converts \(y = ax^2 + bx + c\) to turning-point form (see the page on parabolas by completing the square).
Turning-point form and what each number does to \(y = x^2\):
| Change to the equation | Effect on the graph |
|---|---|
| \(x \to x - h\) | \(h\) units right (left if \(h \lt 0\)) |
| \(+\,k\) on the right | \(k\) units up (down if \(k \lt 0\)) |
| multiply the right by \(a\) | vertical dilation by factor \(|a|\) |
| multiply the right by \(-1\) | reflection in the \(x\)-axis |
Completing the square:
How to sketch \(y = a(x - h)^2 + k\)
- Turning point: plot \((h,\ k)\). Read \(h\) as the value that makes the bracket zero.
- Concavity: up if \(a \gt 0\), down if \(a \lt 0\); larger \(|a|\) means narrower.
- \(y\)-intercept: substitute \(x = 0\).
- \(x\)-intercepts: set \(y = 0\), make \((x - h)^2\) the subject and take \(\pm\) square roots. A negative right-hand side means there are none.
- Sketch a smooth symmetric curve through the labelled points.
Problem types
- Describing transformations: dilate (and reflect), then translate horizontally, then vertically. Track the turning point: \((0,\ 0) \to (h,\ 0) \to (h,\ k)\).
- Writing the equation from transformations: apply each change to the equation in the order given (Example 2).
- Translating \(y = ax^2\): moved \(h\) right and \(k\) up it becomes \(y = a(x - h)^2 + k\). Expand if the question asks for \(y = ax^2 + bx + c\); put \(x = 0\) for the \(y\)-intercept.
- From a graph or a description: turning point and one point \(\to\) \(y = a(x - h)^2 + k\), then solve for \(a\). Two \(x\)-intercepts and one point \(\to\) \(y = a(x - p)(x - q)\). Two points with equal \(y\)-values give the axis \(x = h\).
- Turning point and the \(y\)-intercept: the \(y\)-intercept is the point \((0, c)\); substitute it into \(y = a(x - h)^2 + k\) to find \(a\).
Compare with \(y = a(x - h)^2 + k\): \(a = -2\), \(h = 3\), \(k = 8\). Build the equation in this order:
| \(y = x^2\) | \(\to\) | \(y = 2x^2\ \ \text{(dilate by }2\text{)}\) |
| \(\) | \(\to\) | \(y = -2x^2\ \ \text{(reflect in }x\text{-axis)}\) |
| \(\) | \(\to\) | \(y = -2(x - 3)^2\ \ \text{(3 right)}\) |
| \(\) | \(\to\) | \(y = -2(x - 3)^2 + 8\ \ \text{(8 up)}\) |
Turning point \(\mathbf{(3,\ 8)}\), a maximum. \(y\)-intercept: \(-2(-3)^2 + 8 = \mathbf{-10}\). \(x\)-intercepts:
| \(2(x - 3)^2\) | \(=\) | \(8\) |
| \((x - 3)^2\) | \(=\) | \(4\) |
| \(x - 3\) | \(=\) | \(\pm 2\) |
| \(x\) | \(=\) | \(\mathbf{1}\ \text{or}\ \mathbf{5}\) |
(i) the downward translation is done last
(ii) the downward translation is done before the dilation.
Translate left: replace \(x\) by \(x + 2\). Dilate: multiply the whole right-hand side by \(\dfrac{1}{2}\). Translate down: subtract \(3\) from the whole right-hand side.
(i) Left, dilate, then down:
| \(y\) | \(=\) | \((x + 2)^2\) |
| \(y\) | \(=\) | \(\dfrac{1}{2}(x + 2)^2\) |
| \(y\) | \(=\) | \(\boldsymbol{\dfrac{1}{2}(x + 2)^2 - 3}\) |
(ii) Left, down, then dilate:
| \(y\) | \(=\) | \((x + 2)^2 - 3\) |
| \(y\) | \(=\) | \(\dfrac{1}{2}\left[(x + 2)^2 - 3\right]\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{1}{2}(x + 2)^2 - \dfrac{3}{2}}\) |
The turning points are \((-2,\ -3)\) and \((-2,\ -1.5)\): different parabolas. A dilation done after a vertical translation also halves that translation. Translations left or right are not affected by a vertical dilation.
(ii) Describe the translation that maps \(y = (x + 2)^2 - 3\) onto \(y = (x - 3)^2 + 4\).
(i) \(4\) right: replace \(x\) by \(x - 4\). \(1\) down: subtract \(1\).
| \(y\) | \(=\) | \(3(x - 4)^2 - 1\) |
| \(\) | \(=\) | \(3(x^2 - 8x + 16) - 1\) |
| \(\) | \(=\) | \(\boldsymbol{3x^2 - 24x + 47}\) |
(ii) Both have \(a = 1\), so only the turning point moves.
| \((-2,\ -3)\) | \(\to\) | \((3,\ 4)\) |
| \(\text{across}\) | \(\) | \(3 - (-2) = 5\ \text{right}\) |
| \(\text{up}\) | \(\) | \(4 - (-3) = 7\ \text{up}\) |
The translation is \(\mathbf{5}\) units right and \(\mathbf{7}\) units up.
(i) Use turning-point form \(y = a(x + 2)^2 + 3\) and substitute \((0,\ -5)\):
| \(-5\) | \(=\) | \(a(0 + 2)^2 + 3\) |
| \(4a\) | \(=\) | \(-8\) |
| \(a\) | \(=\) | \(-2\) |
So \(y = \boldsymbol{-2(x + 2)^2 + 3}\). Let \(y = 0\):
| \(2(x + 2)^2\) | \(=\) | \(3\) |
| \((x + 2)^2\) | \(=\) | \(\dfrac{3}{2}\) |
| \(x\) | \(=\) | \(\mathbf{-2 \pm \sqrt{\dfrac{3}{2}}}\) |
\(\sqrt{\dfrac{3}{2}} = \sqrt{1.5} \approx 1.22\), so the \(x\)-intercepts are about \(-3.22\) and \(-0.78\).
Common pitfalls
Frequently asked questions
How do you find the turning point from y = a(x - h)^2 + k?
The turning point is \((h,\ k)\). The \(x\)-coordinate makes the bracket zero, so \(y = 5(x + 1)^2 - 7\) has turning point \((-1,\ -7)\).
Why does (x + 3) squared move the parabola left?
The vertex is where the bracket equals zero, which happens at x = negative 3. Every point of y = x squared now appears 3 units earlier, so the graph shifts left.
What does the a value do to a parabola?
It dilates the parabola vertically. If the size of a is bigger than 1 the parabola is narrower, if it is between 0 and 1 it is wider, and if a is negative the parabola is reflected in the x-axis and opens downwards.
Does the order of transformations matter?
Sometimes. A vertical dilation done after a vertical translation also stretches that translation, so the results differ. Horizontal translations and vertical dilations can be swapped without changing the result.
How do you sketch a parabola without graphing software?
Write it in turning-point form by completing the square, plot the turning point, decide the concavity from a, then find and plot the y-intercept and any x-intercepts before drawing a smooth symmetric curve.