Vertex Form and Transformations

Every parabola is a moved, stretched, or flipped copy of the simplest one, $y = x^2$. When a quadratic is written in vertex form, $y = a(x - h)^2 + k$, the numbers $a$, $h$ and $k$ tell you exactly how it was moved, and the vertex is right there in the equation. That makes vertex form the fastest way to sketch a parabola by hand.

Key ideas

Start with y = x²

Here is a table of values for $y = x^2$:

$x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$
$y$	$9$	$4$	$1$	$0$	$1$	$4$	$9$

Its vertex is $(0, 0)$ and its axis of symmetry is $x = 0$. Look at how the points climb from the vertex: go over 1, up 1, then over 1, up 3, then over 1, up 5. This step pattern ($1, 3, 5$) works on both sides of the vertex, and it’s the key to sketching quickly. (The numbers $1, 3, 5$ are the first differences from the table.)

The effect of k: vertical translation

In $y = x^2 + k$, every $y$-value of $y = x^2$ has $k$ added to it, so the whole graph moves up $k$ units if $k \gt 0$, or down if $k \lt 0$. The vertex moves to $(0, k)$.

For example, $y = x^2 + 3$ is $y = x^2$ moved up $3$, and $y = x^2 - 2$ is $y = x^2$ moved down $2$.

The effect of h: horizontal translation

In $y = (x - h)^2$, the graph moves right $h$ units if $h \gt 0$, or left if $h \lt 0$. The vertex moves to $(h, 0)$.

The sign looks backwards at first, so think about where the vertex is. The vertex is where the bracket equals $0$:

• $y = (x - 2)^2$: the bracket is $0$ when $x = 2$, so the graph moves right $2$ ($h = 2$).
• $y = (x + 3)^2$: the bracket is $0$ when $x = -3$, so the graph moves left $3$. Here $x + 3 = x - (-3)$, so $h = -3$.

The effect of a: stretch, compression, and reflection

In $y = ax^2$, every $y$-value of $y = x^2$ is multiplied by $a$.

• If $a \gt 1$ (or $a \lt -1$), the graph is vertically stretched: it looks narrower. The stretch factor is the size of $a$, ignoring its sign (so $a = -3$ is a stretch by a factor of $3$).
• If $a$ is between $-1$ and $1$ (but not $0$), the graph is vertically compressed: it looks wider.
• If $a \lt 0$, the graph is also reflected in the $x$-axis: it opens down.

The vertex stays at $(0, 0)$, because $a \times 0 = 0$.

Changing one parameter at a time: $k$ moves the graph up or down, $h$ moves it left or right, and $a$ stretches, compresses or reflects it.

Putting it together

In vertex form

$$y = a(x - h)^2 + k$$

all three transformations happen at once:

Parameter	Effect on the graph of $y = x^2$
$a$	vertical stretch if $a \gt 1$ or $a \lt -1$; vertical compression if $a$ is between $-1$ and $1$; if $a \lt 0$, also a reflection in the $x$-axis
$h$	translation right $h$ units ($h \gt 0$) or left ($h \lt 0$)
$k$	translation up $k$ units ($k \gt 0$) or down ($k \lt 0$)

From vertex form you can read:

• the vertex $(h, k)$
• the axis of symmetry $x = h$
• the direction of opening: up if $a \gt 0$, down if $a \lt 0$
• the minimum value $k$ (if $a \gt 0$) or maximum value $k$ (if $a \lt 0$)

Sketching with the step pattern

To sketch $y = a(x - h)^2 + k$ by hand:

1. Plot the vertex $(h, k)$ and draw the axis of symmetry $x = h$ lightly.
2. From the vertex, use the step pattern multiplied by $a$: over $1$, up $1a$; over $1$ more, up $3a$; over $1$ more, up $5a$. (If $a$ is negative, “up” by a negative amount means down.)
3. Mirror those points on the other side of the axis.
4. Join the points with a smooth U-shaped curve.

Finding the equation from a graph

If you know the vertex $(h, k)$ and one other point $(x, y)$ on the parabola:

1. Write $y = a(x - h)^2 + k$ with the vertex filled in.
2. Substitute the other point for $x$ and $y$.
3. Solve for $a$.

You’ll learn how to turn standard form into vertex form later in this unit, in completing the square.

Worked examples

Example 1: Reading vertex form

For $y = -3(x + 2)^2 + 5$, describe the transformations of $y = x^2$, and state the vertex, axis of symmetry, direction of opening, and maximum or minimum value.

Solution. Write the bracket as $x - (-2)$ to see the values: $a = -3$, $h = -2$, $k = 5$.

Transformations of $y = x^2$:

• vertical stretch by a factor of $3$
• reflection in the $x$-axis (because $a$ is negative)
• translation $2$ units left and $5$ units up

Features:

• vertex $(-2, 5)$
• axis of symmetry $x = -2$
• opens down, since $a = -3 \lt 0$
• maximum value $5$

Example 2: Sketching with the step pattern

Sketch $y = 2(x - 1)^2 - 3$.

Solution. Here $a = 2$, $h = 1$, $k = -3$. The vertex is $(1, -3)$ and the axis is $x = 1$.

Multiply the step pattern $1, 3, 5$ by $a = 2$ to get $2, 6, 10$. From the vertex:

• over $1$, up $2$: $(2, -1)$
• over $1$ more, up $6$: $(3, 5)$

Mirror these across $x = 1$: $(0, -1)$ and $(-1, 5)$. Join the points with a smooth curve.

With $a = 2$, the steps are $2$, then $6$ (instead of $1$, then $3$).

Check the $y$-intercept by substituting $x = 0$: $y = 2(0 - 1)^2 - 3 = 2 - 3 = -1$. ✓ That matches the point $(0, -1)$.

Example 3: A parabola that opens down

Sketch $y = -\tfrac{1}{2}(x + 3)^2 + 2$, and find its zeros from the sketch.

Solution. Here $a = -\tfrac{1}{2}$, $h = -3$, $k = 2$. The vertex is $(-3, 2)$ and the axis is $x = -3$.

Multiply the step pattern by $-\tfrac{1}{2}$: the steps are $-\tfrac{1}{2}, -\tfrac{3}{2}, -\tfrac{5}{2}$, so the graph goes down from the vertex.

• over $1$, down $\tfrac{1}{2}$: $(-2, 1.5)$
• over $1$ more, down $\tfrac{3}{2}$: $(-1, 0)$
• mirror points: $(-4, 1.5)$ and $(-5, 0)$

The graph crosses the $x$-axis at $(-1, 0)$ and $(-5, 0)$, so the zeros are $-1$ and $-5$.

Check $x = -1$: $y = -\tfrac{1}{2}(-1 + 3)^2 + 2 = -\tfrac{1}{2}(4) + 2 = 0$. ✓

Example 4: Finding the equation from the vertex and a point

A parabola has its vertex at $(2, -4)$ and passes through $(0, 8)$. Find its equation in vertex form.

Solution. Fill in the vertex: $h = 2$, $k = -4$.

$$y = a(x - 2)^2 - 4$$

Substitute the point $(0, 8)$, so $x = 0$ and $y = 8$:

$$\begin{aligned} 8 &= a(0 - 2)^2 - 4 \\ 8 &= 4a - 4 \\ 12 &= 4a \\ a &= 3 \end{aligned}$$

The equation is $y = 3(x - 2)^2 - 4$.

Check: at $x = 0$, $y = 3(4) - 4 = 8$. ✓

Common mistakes

Getting the sign of $h$ backwards. $y = (x + 3)^2$ moves left $3$, and its vertex is $(-3, 0)$, not $(3, 0)$. Ask yourself: what value of $x$ makes the bracket zero?

Using the plain step pattern when $a \ne 1$. For $y = 2(x - 1)^2 - 3$, the steps are $2, 6, 10$, not $1, 3, 5$. Always multiply by $a$.

Stepping up when $a$ is negative. If $a \lt 0$, the parabola opens down, so the steps go down from the vertex.

Thinking a bigger $a$ makes a wider parabola. It’s the opposite: a vertical stretch ($a = 3$) makes the graph narrower, and a compression ($a = \tfrac{1}{3}$) makes it wider.

Mixing up the two translations. The number inside the bracket ($h$) moves the graph sideways. The number outside ($k$) moves it up or down.

Solving for $a$ with the vertex instead of the other point. Substituting the vertex itself always gives $k = k$, which tells you nothing. Use a different point on the graph.

Practice

1. (Warm-up) State the vertex and axis of symmetry of each parabola.

• (a) $y = (x - 4)^2 + 1$
• (b) $y = -2(x + 5)^2$
• (c) $y = 0.5x^2 - 7$

Solution

(a) Vertex $(4, 1)$, axis $x = 4$.

(b) Vertex $(-5, 0)$, axis $x = -5$. (There’s no $k$ written, so $k = 0$.)

(c) Vertex $(0, -7)$, axis $x = 0$. (There’s no bracket, so $h = 0$.)

2. (Warm-up) Describe how the graph of $y = (x + 6)^2 - 2$ is related to the graph of $y = x^2$.

Solution

$h = -6$ and $k = -2$: it is the graph of $y = x^2$ translated $6$ units left and $2$ units down. The vertex moves from $(0, 0)$ to $(-6, -2)$.

3. (Warm-up) The graph of $y = x^2$ is reflected in the $x$-axis, then translated $3$ units right and $1$ unit up. Write the equation of the new parabola.

Solution

A reflection in the $x$-axis gives $a = -1$; right $3$ gives $h = 3$; up $1$ gives $k = 1$.

$$y = -(x - 3)^2 + 1$$

4. (Core) Sketch $y = -(x - 2)^2 + 9$ using the step pattern. Label the vertex, the $y$-intercept, and the zeros.

Solution

The vertex is $(2, 9)$ and $a = -1$, so the steps are $-1, -3, -5$ (going down).

• over $1$, down $1$: $(3, 8)$; mirror point $(1, 8)$
• over $1$ more, down $3$: $(4, 5)$; mirror point $(0, 5)$
• over $1$ more, down $5$: $(5, 0)$; mirror point $(-1, 0)$

The $y$-intercept is $5$ (the point $(0, 5)$). The zeros are $-1$ and $5$.

Check: $y = -(5 - 2)^2 + 9 = -9 + 9 = 0$. ✓

5. (Core) Sketch $y = 0.5(x + 4)^2 - 2$. Find the zeros and the $y$-intercept.

Solution

The vertex is $(-4, -2)$ and $a = 0.5$, so the steps are $0.5, 1.5, 2.5$.

• over $1$, up $0.5$: $(-3, -1.5)$; mirror point $(-5, -1.5)$
• over $1$ more, up $1.5$: $(-2, 0)$; mirror point $(-6, 0)$
• over $1$ more, up $2.5$: $(-1, 2.5)$; mirror point $(-7, 2.5)$

The zeros are $-6$ and $-2$.

For the $y$-intercept, substitute $x = 0$: $y = 0.5(0 + 4)^2 - 2 = 0.5(16) - 2 = 6$.

6. (Core) A parabola has its vertex at $(-1, 5)$ and passes through $(1, -3)$. Find its equation in vertex form.

Solution

$$y = a(x + 1)^2 + 5$$

Substitute $(1, -3)$:

$$\begin{aligned} -3 &= a(1 + 1)^2 + 5 \\ -3 &= 4a + 5 \\ -8 &= 4a \\ a &= -2 \end{aligned}$$

The equation is $y = -2(x + 1)^2 + 5$.

7. (Core) The entrance to a tunnel is shaped like a parabola. It is $6$ m high in the middle and $8$ m wide at the ground. Place the origin on the ground directly below the highest point.

• (a) Find an equation for the arch in vertex form.
• (b) How high is the arch $2$ m from the centre line?

Solution

(a) The vertex is $(0, 6)$. The arch is $8$ m wide, so it meets the ground $4$ m on each side: at $(4, 0)$ and $(-4, 0)$.

$$y = a(x - 0)^2 + 6 = ax^2 + 6$$

Substitute $(4, 0)$:

$$0 = 16a + 6 \quad\Rightarrow\quad a = -\frac{6}{16} = -\frac{3}{8}$$

The equation is $y = -\tfrac{3}{8}x^2 + 6$.

(b) Substitute $x = 2$:

$$y = -\frac{3}{8}(2)^2 + 6 = -\frac{3}{8}(4) + 6 = -1.5 + 6 = 4.5$$

The arch is $4.5$ m high, $2$ m from the centre.

8. (Challenge) A parabola has zeros at $1$ and $7$ and a maximum value of $18$. Find its equation in vertex form.

Solution

The axis of symmetry is halfway between the zeros: $x = \dfrac{1 + 7}{2} = 4$. The maximum value is $18$, so the vertex is $(4, 18)$.

$$y = a(x - 4)^2 + 18$$

Substitute the zero $(1, 0)$:

$$0 = a(1 - 4)^2 + 18 = 9a + 18 \quad\Rightarrow\quad a = -2$$

The equation is $y = -2(x - 4)^2 + 18$. Check the other zero: $-2(7 - 4)^2 + 18 = -18 + 18 = 0$. ✓

9. (Challenge) The point $(3, 9)$ is on the graph of $y = x^2$. Where does this point end up on the graph of $y = 2(x + 1)^2 - 5$? Check your answer by substituting.

Solution

Apply the transformations to the point in order:

• vertical stretch by $2$: the $y$-coordinate doubles, giving $(3, 18)$
• left $1$ and down $5$: giving $(3 - 1,\ 18 - 5) = (2, 13)$

So $(3, 9)$ moves to $(2, 13)$. In mapping notation, $(x, y) \to (x - 1,\ 2y - 5)$.

Check: $y = 2(2 + 1)^2 - 5 = 2(9) - 5 = 13$. ✓

Notice the order: the stretch comes before the vertical translation. Doing it the other way would give $2(9 - 5) = 8$, which is wrong.