Sketching Parabolas Using Transformations
Any parabola can be drawn by transforming \(y=x^2\). In the turning-point form \(y=a(x-h)^2+k\), the vertex is \((h,k)\), the value of \(a\) stretches the graph and a negative \(a\) reflects it in the \(x\)-axis. In NSW Year 10 Advanced maths you complete the square to reach this form and sketch parabolas from it (NESA outcome MA5‑NLI‑P‑01).
You will describe the transformations from \(y=x^2\) to graphs such as \(y=3(x-1)^2+2\), write equations after reflections and translations, and complete the square to find the turning point and the number of \(x\)-intercepts. You will sketch parabolas showing all intercepts, find equations from a turning point and another point. Harder questions find the path traced by a family of turning points. It is part of Year 10 Advanced Non Linear Relationships Functions and Graphs.
Every question with a fully worked solution.
- Sketching 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 vertex is \((h,\ k)\), and \(a\) dilates the curve and, if negative, reflects it. This Year 10 Advanced maths page shows how to describe the transformations, complete the square to reach this form, sketch without graphing software, and work backwards from a graph 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 \(a \lt 0\).
Completing the square converts \(y = ax^2 + bx + c\) to turning-point form.
It comes up if you are heading for Advanced or Extension maths in Years 11 and 12.
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).
- Completing the square: for \(a \ne 1\), take out \(a\) from the \(x^2\) and \(x\) terms; whatever you add inside the bracket is multiplied by \(a\) outside.
- 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 \(x = h\).
- Unknown coefficients: expand the turning-point form and match it with \(y = x^2 + bx + c\), term by term.
- Families of parabolas: with a letter such as \(m\) in the equation, complete the square to find the turning point in terms of \(m\), then apply the condition given (passes through a point, lies on a line).
Compare with \(y = a(x - h)^2 + k\): \(a = -3\), \(h = 1\), \(k = 12\). Build the equation in this order:
| \(y = x^2\) | \(\to\) | \(y = 3x^2\ \ \text{(dilate by }3\text{)}\) |
| \(\) | \(\to\) | \(y = -3x^2\ \ \text{(reflect in }x\text{-axis)}\) |
| \(\) | \(\to\) | \(y = -3(x - 1)^2\ \ \text{(1 right)}\) |
| \(\) | \(\to\) | \(y = -3(x - 1)^2 + 12\ \ \text{(12 up)}\) |
Turning point \((1,\ 12)\), a maximum (concave down). \(y\)-intercept: \(-3(1) + 12 = 9\). \(x\)-intercepts:
| \(3(x - 1)^2\) | \(=\) | \(12\) |
| \((x - 1)^2\) | \(=\) | \(4\) |
| \(x - 1\) | \(=\) | \(\pm 2\) |
| \(x\) | \(=\) | \(-1\ \text{or}\ 3\) |
(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.
Take out \(-1\) from the \(x\) terms, then complete the square inside the bracket:
| \(y\) | \(=\) | \(-(x^2 - 8x) - 12\) |
| \(\) | \(=\) | \(-(x^2 - 8x + 16 - 16) - 12\) |
| \(\) | \(=\) | \(-(x - 4)^2 + 16 - 12\) |
| \(\) | \(=\) | \(-(x - 4)^2 + 4\) |
This is \(y = x^2\) reflected in the \(x\)-axis, moved \(4\) right and \(4\) up. Turning point \((4,\ 4)\), a maximum. \(y\)-intercept \(-12\).
| \((x - 4)^2\) | \(=\) | \(4\) |
| \(x\) | \(=\) | \(4 \pm 2 = 2\ \text{or}\ 6\) |
Use turning-point form \(y = a(x - 1)^2 - 4\) and substitute \((3,\ 4)\):
| \(4\) | \(=\) | \(a(3 - 1)^2 - 4\) |
| \(4a\) | \(=\) | \(8\) |
| \(a\) | \(=\) | \(2\) |
So \(y = \boldsymbol{2(x - 1)^2 - 4}\). Let \(y = 0\):
| \(2(x - 1)^2\) | \(=\) | \(4\) |
| \((x - 1)^2\) | \(=\) | \(2\) |
| \(x\) | \(=\) | \(\mathbf{1 \pm \sqrt{2}}\) |
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.