MPM2D Transformations and Vertex Form: Notes and Practice

A fountain designer wants six identical arcs of water, each starting from a different nozzle around a pool. Every arc has the same shape; only its position changes. Another fountain needs taller, narrower jets, and a third shoots its water downward into a basin. All of these are the same basic curve, the parabola $y = x^{2}$ , after it has been moved, stretched or flipped. Once you know how each change shows up in an equation, you can sketch a parabola in seconds without a table of values, and you can go the other way: look at a graph and write its equation. The equation that makes all of this visible is called vertex form, $y = a (x - h)^{2} + k$ .

Start with the parent parabola $y = x^{2}$ . Its table is $x = - 3, - 2, - 1, 0, 1, 2, 3$ with $y = 9, 4, 1, 0, 1, 4, 9$ . The vertex is $(0, 0)$ and the axis of symmetry is $x = 0$ . Notice the step pattern: starting at the vertex, go 1 unit right and 1 unit up to reach $(1, 1)$ ; go 1 more right and 3 up to reach $(2, 4)$ ; then 1 more right and 5 up to reach $(3, 9)$ . The same steps work to the left. This 1, 3, 5 pattern (the first differences of $y = x^{2}$ ) lets you sketch any parabola from its vertex. Now change the equation one parameter at a time.

Vertical translation, $y = x^{2} + k$ : every $y$ -value is $k$ more, so the whole graph slides up $k$ units if $k > 0$ or down if $k < 0$ . The vertex moves to $(0, k)$ . Example: $y = x^{2} - 2$ is $y = x^{2}$ moved down 2, vertex $(0, - 2)$ .
Horizontal translation, $y = (x - h)^{2}$ : the graph slides right $h$ units if $h > 0$ or left if $h < 0$ . The vertex moves to $(h, 0)$ . Example: $y = (x - 3)^{2}$ moves right 3, because the bracket becomes $0$ (the lowest output) when $x = 3$ . And $y = (x + 2)^{2} = (x - (- 2))^{2}$ moves left 2, with vertex $(- 2, 0)$ .
Vertical stretch or compression, $y = a x^{2}$ with $a > 0$ : every $y$ -value is multiplied by $a$ . If $a > 1$ the graph is stretched vertically and looks narrower; if $0 < a < 1$ it is compressed vertically and looks wider. The step pattern becomes $a, 3 a, 5 a$ : for $y = 2 x^{2}$ the steps are 2, 6, 10.
Reflection in the x-axis, $y = - x^{2}$ : every $y$ -value changes sign, so the parabola flips and opens down. With $a < 0$ the graph is reflected, and it is also stretched if $a < - 1$ or compressed if $- 1 < a < 0$ . For $y = - \frac{1}{2} x^{2}$ the steps are $- \frac{1}{2}, - \frac{3}{2}, - \frac{5}{2}$ .

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Choose the parent x² and drag the sliders. The sim uses different letters: its a is the a of this lesson, its d slider (horizontal shift) plays the role of h, and its c slider (vertical shift) plays the role of k. Keep the sim's k slider at 1, because that slider is a horizontal stretch, which this lesson does not use.

Two arcing jets of water from nozzles at the edge of a stone fountain pool — PhotoThese two jets have the same shape but start from different nozzles, so one arc is a translation of the other, just like shifting y equals x squared.GuidoB, CC BY-SA 3.0, via Wikimedia Commons

Putting it together: vertex form

y = a (x - h)^{2} + k vertex (h, k), axis of symmetry x = h

Parameter	What it does	Example
$a > 0$	opens up; vertex is a minimum	$y = 3 (x - 1)^{2} + 2$ : minimum value $2$
$a < 0$	reflection in the x-axis; opens down; vertex is a maximum	$y = - (x - 1)^{2} + 2$ : maximum value $2$
$a > 1$ or $a < - 1$	vertical stretch (narrower); a < −1 also reflects	$y = 4 x^{2}$ : steps $4, 12, 20$
$- 1 < a < 1$ , $a \neq = 0$	vertical compression (wider); a < 0 also reflects	$y = 0.5 x^{2}$ : steps $0.5, 1.5, 2.5$
$h$	horizontal translation: right if h > 0, left if h < 0	$y = (x + 5)^{2}$ : left 5, $h = - 5$
$k$	vertical translation: up if k > 0, down if k < 0	$y = x^{2} - 6$ : down 6

The roles of a, h and k in y = a(x − h)² + k

Fig 1Building y = −½(x − 3)² + 4 from y = x². The dashed parent is first reflected in the x-axis and compressed by a factor of one half (dotted), then translated 3 right and 4 up (solid). The marked points come from the step pattern −½, −3/2, −5/2.

Read $a$ , $h$ and $k$ . Watch the sign of $h$ : in $y = 2 (x + 1)^{2} - 8$ , $h = - 1$ and $k = - 8$ .
Plot the vertex $(h, k)$ and draw the axis of symmetry $x = h$ as a dashed line.
From the vertex, step 1 unit right and 1 unit left and move $a$ units vertically (down if $a$ is negative). Then step 1 more and move $3 a$ , then 1 more and move $5 a$ .
Join the points with a smooth U-shaped curve (not a V), and label the vertex and the equation.

Why does this order work? Each point $(x, y)$ on $y = x^{2}$ goes to $(x + h, a y + k)$ on $y = a (x - h)^{2} + k$ : first multiply the $y$ -coordinate by $a$ (stretch, compress or reflect), then add $h$ to $x$ and $k$ to $y$ (translate). For example, the point $(1, 1)$ on $y = x^{2}$ goes to $(1 + 3, - \frac{1}{2} (1) + 4) = (4, 3.5)$ on $y = - \frac{1}{2} (x - 3)^{2} + 4$ . If you translated first and then multiplied by $a$ , you would also multiply $k$ by $a$ and put the vertex in the wrong place.

Count the zeros without solving

Vertex form tells you how many zeros there are. If

a > 0

and

k > 0

, the lowest point is above the x-axis: no zeros. If

a < 0

and

k < 0

, the highest point is below it: no zeros. If

k = 0

, the vertex is on the x-axis: one zero. If

a

and

k

have opposite signs, the parabola crosses the x-axis twice.

VideoSee how a, h and k in vertex form stretch, reflect and translate the parent parabola y equals x squared.Watch on YouTube

Finding the equation from a graph

To write the equation of a parabola you are shown, read the vertex $(h, k)$ from the graph and substitute it into $y = a (x - h)^{2} + k$ . The vertex alone does not fix the shape, so pick one more point $(x, y)$ on the curve whose coordinates you can read exactly (a grid point), substitute it, and solve for $a$ . Then check: does the sign of $a$ match the direction of opening? Does the step pattern from the vertex land on the graph? The Parabola Forms Lab below lets you test your answer by dragging the vertex and editing $a$ .

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Edit a, h and k in the vertex form panel or drag the vertex on the graph. Tick the box that shows y = x² and the transformations to compare your parabola with the parent.

A concrete arch bridge spanning a rocky river gorge with green and autumn trees — PhotoAn arch like the one under this bridge in New Zealand can be modelled with a parabola: with the vertex at the top and one point where it meets the rock, you could write its equation in vertex form.Bernard Spragg. NZ from Christchurch, New Zealand, CC0, via Wikimedia Commons

Transformations and Vertex Form

Key things to remember

Putting it together: vertex form

Finding the equation from a graph

Worked examples

Where marks get lost

Flashcards 8

Practice