VertexMath · MPM2D · Free worksheet

Quadratic Functions in Vertex Form Review

Try each problem first. You don’t need to finish everything.

Core path ~55 min Full sheet ~1 h 50 min

Recommended: K1–K5, then one A and one T problem

K = Knowledge & Understanding · A = Application · T = Thinking · Boss = Challenge

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Quick reference

Vertex form $y=a(x-h)^2+k$ : the vertex is $(h,k)$ and the axis of symmetry is $x=h$ . In $(x+3)^2$ , $h=-3$ ; the $k$ keeps its sign.
Parameter $a$ : $|a|>1$ stretches the parabola (narrower), $0<|a|<1$ compresses it (wider), and $a<0$ reflects it in the $x$ -axis (opens down, maximum).
Square first, then multiply: for $y=3x^2$ at $x=2$ , $y=3(2)^2=12$ , not $(3\cdot 2)^2=36$ .
Finding $a$ : substitute the vertex for $h$ and $k$ , a second point for $x$ and $y$ , then solve. Remember to square $(x-h)$ .
Completing the square: factor $a$ out of the $x^2$ and $x$ terms only. Whatever you add inside the bracket is multiplied by $a$ , so subtract $a$ times it outside.

Show your work on separate paper. Show your work in the dot-grid space.

Core Skills (K)

Stretches & Reflections · Translations · Vertex & a Point · Completing the Square · Three Forms

K1 The Role of $a$

a)
Complete the table of values for $y=3x^2$ .

$x$ $-2$ $-1$ $0$ $1$ $2$
$y$
b)
The parabola $y=ax^2$ is graphed below. Determine $a$ . Each grid square is 1 unit.
c)
Order from narrowest to widest: $y=-4x^2$ , $y=0.5x^2$ , $y=2x^2$ , $y=-0.25x^2$ .

$x$	$-2$	$-1$	$0$	$1$	$2$
$y$

Stuck? Peek at a hint

Square $x$ first, then multiply by $a$ . For (b), substitute a point on the curve other than the vertex into $y=ax^2$ .

K2 Reading the Transformations

a)
State the vertex and the axis of symmetry of $y=(x+3)^2-4$ .
b)
For $y=-2(x-5)^2+1$ , state the vertex, the direction of opening and the $y$ -intercept.
c)
The parabola $y=x^2$ is stretched vertically by a factor of $3$ , then translated $4$ units left and $2$ units down. Write its equation.

Stuck? Peek at a hint

Write the bracket as $(x-h)$ : $(x+3)$ is $(x-(-3))$ . The $y$ -intercept is the value of $y$ when $x=0$ .

K3 Vertex and One Point

Determine the equation of each parabola in vertex form.

a)
Vertex $(2,3)$ , passing through $(4,11)$ .
b)
The parabola graphed below. Each grid square is 1 unit.
c)
Vertex $(-4,0)$ and $y$ -intercept $8$ .

Stuck? Peek at a hint

Substitute the vertex for $h$ and $k$ and the other point for $x$ and $y$ , then solve for $a$ .

K4 Completing the Square

Write each relation in vertex form by completing the square. Then state the vertex.

a)
$y=x^2+8x+11$
b)
$y=3x^2-18x+25$
c)
$y=-2x^2-12x-13$

Stuck? Peek at a hint

Factor $a$ out of the $x^2$ and $x$ terms only. Inside the bracket, add and subtract the square of half the coefficient of $x$ .

K5 Three Forms, One Parabola

a)
Factor $y=x^2-2x-15$ and state its zeros.
b)
Use the zeros to determine the vertex of $y=x^2-2x-15$ . Then write the relation in vertex form.
c)
Write $y=2(x-3)^2-8$ in standard form and in factored form.

Stuck? Peek at a hint

For (a), find two numbers with product $-15$ and sum $-2$ . The axis of symmetry is halfway between the zeros.

Real World (A)

Quadratic Models · Choosing the Best Form

A1 Gym Ceiling

In a school gym, a volleyball is bumped from a height of $2$ m. Its height, in metres, after $t$ seconds is $h(t)=-5(t-1)^2+7$ .

a)
State the maximum height of the ball and when it occurs.
b)
A teammate's hands are $2$ m above the floor. When does the ball reach them on the way down?
c)
A harder bump from the same height is modelled by $g(t)=-5(t-1.2)^2+9.2$ . The gym ceiling is $9$ m high. State the maximum height the model predicts, then say what really happens to the ball.

Stuck? Peek at a hint

Read the maximum from the vertex. The parabola is symmetric about $t=1$ , and the ball starts at $2$ m when $t=0$ .

A2 Ski Hill Day Passes

A ski hill near Collingwood finds that its daily revenue from day passes is greatest, $14 400, at a price of $60 per pass. At a price of $40, the daily revenue is $12 800.

a)
Write a model $R(p)$ in vertex form for the daily revenue, in dollars, at a price of $p$ dollars per pass.
b)
Use your model to predict the daily revenue at $75 per pass.
c)
Use your model to predict the daily revenue at $130 per pass. What does the result mean for the ski hill?

Stuck? Peek at a hint

The vertex is $(60,\,14\,400)$ . Substitute the point $(40,\,12\,800)$ to find $a$ .

A3 Parade Arch

A parabolic arch frames the entrance to a fairground in London, Ontario. Its height, in metres, is $h(x)=-0.5x^2+6x-10$ , where $x$ is the horizontal distance, in metres, from a lamp post.

a)
Write $h(x)$ in vertex form. State the maximum height of the arch and how far from the lamp post it occurs.
b)
Write $h(x)$ in factored form. How wide is the arch at ground level?
c)
A parade float has a flat top. It is $4.4$ m wide and $6$ m tall, and it drives through the centre of the arch. Does it fit?

Stuck? Peek at a hint

For vertex form, factor $-0.5$ out of the $x^2$ and $x$ terms and complete the square. For factored form, factor $-0.5$ out of all three terms.

Think It Through (T)

Error Analysis · Building Parabolas · Counterexample

T1 Spot the Flaw

A student completes the square for $y=2x^2+12x+5$ :
" $y=2(x^2+6x)+5=2(x^2+6x+9)+5-9=2(x+3)^2-4$ . The minimum value is $-4$ , at $x=-3$ ."

a)
Evaluate the original relation at $x=-3$ . What does the result tell you about the student's answer?
b)
Identify the error and write the correct vertex form.
c)
Without completing the square again, write $y=2x^2+12x+20$ in vertex form. Explain your reasoning.

Stuck? Peek at a hint

In (a), substitute $x=-3$ into $y=2x^2+12x+5$ , not into the student's answer.

T2 Build It Backwards

Use vertex form $y=a(x-h)^2+k$ in each part.

a)
Write an equation of a parabola with vertex $(-2,5)$ that opens down and is narrower than $y=x^2$ . Many answers are possible.
b)
A parabola has axis of symmetry $x=3$ and passes through $(1,0)$ and $(0,-10)$ . Determine its equation.
c)
A student claims: "Every parabola that opens down crosses the $x$ -axis." Give a counterexample. Then state the condition on $k$ for $y=a(x-h)^2+k$ , with $a<0$ , to have two zeros.

Stuck? Peek at a hint

In (b), the axis gives $h$ ; substitute each point to get two equations in $a$ and $k$ .

Boss Level

Optional challenge

Closest Point on a Line

The point $P(2,7)$ lies off the line $y=2x$ . Every point on the line has the form $Q(x,\,2x)$ .

a)
Write $PQ^2$ as a quadratic in $x$ , in standard form.
b)
Complete the square to determine the point $Q$ closest to $P$ , and the shortest distance from $P$ to the line.
c)
The point on the line straight across from $P$ is $(3.5,7)$ . Determine its distance from $P$ . Then show that $PQ$ is perpendicular to the line.

Stuck? Peek at a hint

Use the distance formula with $Q(x,\,2x)$ and keep $PQ^2$ without the root: the smallest $PQ^2$ gives the smallest $PQ$ .

Answer key

Final answers only. Try first, then check.

Core Skills

a) $12,\ 3,\ 0,\ 3,\ 12$
b) $a=-\frac{1}{2}$
c) $y=-4x^2$ , $y=2x^2$ , $y=0.5x^2$ , $y=-0.25x^2$

a) Vertex $(-3,-4)$ ; axis $x=-3$
b) Vertex $(5,1)$ ; opens down; $y$ -intercept $-49$
c) $y=3(x+4)^2-2$

a) $y=2(x-2)^2+3$
b) $y=-2(x+1)^2+4$
c) $y=\frac{1}{2}(x+4)^2$

a) $y=(x+4)^2-5$ ; vertex $(-4,-5)$
b) $y=3(x-3)^2-2$ ; vertex $(3,-2)$
c) $y=-2(x+3)^2+5$ ; vertex $(-3,5)$

a) $y=(x-5)(x+3)$ ; zeros $5$ and $-3$
b) Vertex $(1,-16)$ ; $y=(x-1)^2-16$
c) $y=2x^2-12x+10$ ; $y=2(x-1)(x-5)$

Real World

a) $7$ m, at $t=1$ s
b) $t=2$ s
c) The model predicts $9.2$ m at $t=1.2$ s, but the ball hits the $9$ m ceiling at $t=1$ s ( $g(1)=9$ ), so it never reaches $9.2$ m and the model stops applying after $t=1$ s.

a) $R(p)=-4(p-60)^2+14\,400$
b) $13 500
c) $R(130)=-5200$ : revenue can't be negative. The model reaches $0$ at $120, so above $120 no passes are sold, the revenue is $0, and the model no longer applies.

a) $h(x)=-0.5(x-6)^2+8$ ; $8$ m, at $6$ m from the lamp post
b) $h(x)=-0.5(x-2)(x-10)$ ; $8$ m wide
c) No: the arch is $8$ m high at its centre, but at the float's edges ( $x=3.8$ and $x=8.2$ ) it is only $5.58$ m high, lower than the $6$ m top.

Think It Through

a) $y=-13$ , not $-4$ : the axis $x=-3$ is right, but the minimum value is wrong.
b) Adding $9$ inside the bracket adds $2\cdot 9=18$ , so $18$ must be subtracted: $y=2(x+3)^2-13$ .
c) $y=2(x+3)^2+2$ : the constant is $15$ more, so the parabola is translated $15$ units up.

a) e.g. $y=-3(x+2)^2+5$ (any $a<-1$ )
b) $y=-2(x-3)^2+8$
c) e.g. $y=-(x-1)^2-2$ (maximum $-2$ , no zeros); two zeros when $k>0$

Boss Level

Boss

a) $PQ^2=5x^2-32x+53$
b) $PQ^2=5(x-3.2)^2+1.8$ ; $Q(3.2,\,6.4)$ ; shortest distance $\sqrt{1.8}\approx 1.34$
c) $1.5$ , longer than $1.34$ ; slope of $PQ=-\frac{1}{2}$ , and $-\frac{1}{2}\times 2=-1$

Free cheat sheet

Quadratics in Vertex Form Cheat Sheet

The whole unit on one page: every rule, a worked example for each idea, and the traps that cost marks.

Unit Summary

One printable page with every rule.

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Build your own

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Phone card

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Most testedWrite $y=2x^2-12x+13$ in vertex form. State the vertex and the minimum value. (card 5)

1The parameter $a$ in $y=ax^2$

$\|a\|>1$	stretch: narrower
$0<\|a\|<1$	compression: wider
$a<0$	reflection: opens down

Trap $y=2x^2$ at $x=3$ : $2(3)^2=18$ , not $(2\cdot3)^2=36$ . Square first.

Trap $y=-0.5x^2$ is a vertical compression, wider than $y=x^2$ : the sign only flips it.

2Vertex form

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

UseVertex $(h,k)$ , axis $x=h$ . Shift right $h$ , up $k$ . $a>0$ : minimum $k$ ; $a<0$ : maximum $k$ .

Example $y=-2(x+3)^2+5$ : vertex $(-3,5)$ , maximum $5$ ; $y$ -intercept $-2(9)+5=-13$ .

Trap $y=(x+4)^2-1$ has vertex $(-4,-1)$ : only $h$ changes sign, $k$ keeps its sign.

TrapKeep $a$ for the $y$ -intercept: $y=-3(x-2)^2+1$ gives $-3(4)+1=-11$ , not $5$ .

3Finding the equation

UsePut the vertex in for $h$ and $k$ , a second point in for $x$ and $y$ , then solve for $a$ .

ExampleVertex $(1,-3)$ , through $(3,5)$ : $5=a(3-1)^2-3$ , so $4a=8$ and $a=2$ .

TrapSquare the bracket: $5=a(2)^2-3$ gives $a=2$ . Without the square, you get $a=4$ .

4Models

UseThe maximum or minimum is $k$ , reached at $x=h$ . Equal heights sit the same distance either side of $x=h$ .

Example $h=-5(t-2)^2+21$ : maximum $21$ m at $t=2$ s; the height at $t=1$ equals the height at $t=3$ .

TrapFar from the vertex a revenue model can go negative. Use a model only where it makes sense.

5Completing the square

UseFactor $a$ from the $x^2$ and $x$ terms. Inside, add (half the $x$ -coefficient) $^2$ ; outside, subtract $a$ times that number.

Example $3x^2+12x+7=3(x^2+4x+4)+7-12=3(x+2)^2-5$ .

TrapAdding $4$ inside $3(\ldots)$ adds $12$ : subtract $12$ outside, not $4$ .

TrapFactoring out a negative flips the $x$ -term: $-2x^2+8x=-2(x^2-4x)$ .

6Three forms

$ax^2+bx+c$	$y$ -intercept $c$
$a(x-r)(x-s)$	zeros $r$ , $s$
$a(x-h)^2+k$	vertex $(h,k)$

Example $x^2-6x+5=(x-1)(x-5)$ : zeros $1$ and $5$ , axis $x=3$ , $y(3)=-4$ , so $(x-3)^2-4$ .

Trap $(x-3)^2=x^2-6x+9$ , not $x^2+9$ .

Which form?

You need the zeros Factored form

You need the max or min Vertex form

You need the starting value Standard form

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Core Skills (K)

K1 The Role of aaa

K2 Reading the Transformations

K3 Vertex and One Point

K4 Completing the Square

K5 Three Forms, One Parabola

Real World (A)

A1 Gym Ceiling

A2 Ski Hill Day Passes

A3 Parade Arch

Think It Through (T)

T1 Spot the Flaw

T2 Build It Backwards

Boss Level

Closest Point on a Line

Answer key

Core Skills

Real World

Think It Through

Boss Level

Quadratics in Vertex Form Cheat Sheet

Unit Summary

Build your own

Phone card

1The parameter aaa in y=ax2y=ax^2y=ax2

2Vertex form

3Finding the equation

4Models

5Completing the square

6Three forms

Which form?

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What is VertexMath?

K1 The Role of $a$

1The parameter $a$ in $y=ax^2$