Parabolas Using Transformations
Every question with a fully worked solution.
- Parabolas Using Transformations - Video - Graphing Quadratic Functions Using Transformations Watch
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 10A maths page shows how to describe the transformations, use function notation, sketch a parabola with its intercepts, and work backwards from a graph or a description to the equation, including a turning point that lies on a given line.
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 \(a \lt 0\).
In function notation, if \(f(x) = x^2\) then \(y = f(x - h) + k\) is the same graph moved \(h\) units right and \(k\) units up: \(y = (x - h)^2 + k\).
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).
- Function notation: \(y = f(x - h) + k\) is \(y = f(x)\) moved \(h\) right and \(k\) up. To write it out, replace every \(x\) in \(f\) by \((x - h)\), then add \(k\).
- 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 \(a\).
- Turning point on a line: if the turning point \((h, k)\) lies on \(y = mx + c\), then \(k = mh + c\). Substitute this and the given point into \(y = (x - h)^2 + k\) and solve the quadratic in \(h\) (Example 4).
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 \(1\).
| \(y\) | \(=\) | \(f(x - 4) - 1\) |
| \(\) | \(=\) | \(3(x - 4)^2 - 1\) |
| \(\) | \(=\) | \(3(x^2 - 8x + 16) - 1\) |
| \(\) | \(=\) | \(\boldsymbol{3x^2 - 24x + 47}\) |
(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.
(ii) The parabola \(y = (x - h)^2 + k\) has its turning point on the line \(y = 2x\) and passes through \((0,\ 8)\). Find the two possible equations.
(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 \(y = 0\):
| \(2(x + 2)^2\) | \(=\) | \(3\) |
| \((x + 2)^2\) | \(=\) | \(\dfrac{3}{2}\) |
| \(x\) | \(=\) | \(-2 \pm \sqrt{\dfrac{3}{2}} = \mathbf{-2 \pm \dfrac{\sqrt{6}}{2}}\) |
\(\sqrt{\dfrac{3}{2}} = \dfrac{\sqrt{3}}{\sqrt{2}} = \dfrac{\sqrt{6}}{2}\), about \(1.22\), so the \(x\)-intercepts are about \(-3.22\) and \(-0.78\).
(ii) The turning point \((h, k)\) is on \(y = 2x\), so \(k = 2h\). Substitute \((0, 8)\):
| \(8\) | \(=\) | \((0 - h)^2 + 2h\) |
| \(h^2 + 2h - 8\) | \(=\) | \(0\) |
| \((h + 4)(h - 2)\) | \(=\) | \(0\) |
\(h = 2\) gives \(k = 4\); \(h = -4\) gives \(k = -8\). The parabolas are \(\boldsymbol{y = (x - 2)^2 + 4}\) and \(\boldsymbol{y = (x + 4)^2 - 8}\). Check \((0, 8)\): \(4 + 4 = 8\) and \(16 - 8 = 8\).
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.