VertexMath · MCR3U · Free worksheet

Quadratic Functions 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

Three forms: vertex $y=a(x-h)^2+k$ shows the vertex $(h,k)$ ; factored $y=a(x-r)(x-s)$ shows the zeros $r$ , $s$ and the axis $x=\frac{r+s}{2}$ ; standard $y=ax^2+bx+c$ shows the $y$ -intercept $c$ .
Completing the square: factor $a$ from the $x^2$ and $x$ terms only. The number subtracted inside the bracket is multiplied by $a$ when it comes out.
Quadratic formula: $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ . The discriminant $b^2-4ac$ is positive (two real roots), zero (one) or negative (none).
Line and parabola: set the two equations equal and collect every term on one side. The discriminant of that equation counts the intersection points.
Radicals: $\sqrt{ab}=\sqrt{a}\sqrt{b}$ , but $\sqrt{a+b}\ne\sqrt{a}+\sqrt{b}$ and $(a+\sqrt{b})^2\ne a^2+b$ .

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

Core Skills (K)

Three Forms · Max & Min · Radicals · Solving · Families & Systems

K1 Three Forms

a)
For $y=-2(x+3)^2+5$ , state the vertex, the axis of symmetry and the direction of opening.
b)
Determine the $y$ -intercept of $y=-2(x+3)^2+5$ .
c)
For $y=3(x-1)(x+5)$ , state the zeros, the axis of symmetry and the vertex.

Stuck? Peek at a hint

Vertex form has $x-h$ inside the bracket, so match each bracket to that pattern. For a $y$ -intercept, substitute $x=0$ .

K2 Completing the Square

Write each function in vertex form. Then state its maximum or minimum value and the value of $x$ where it occurs.

a)
$y=x^2-10x+18$
b)
$y=2x^2+12x+7$
c)
$y=-3x^2+12x-5$

Stuck? Peek at a hint

Factor $a$ from the $x^2$ and $x$ terms only, then add and subtract the square of half the $x$ -coefficient inside the bracket.

K3 Radicals

Simplify fully. Leave no radical in a denominator.

a)
$\sqrt{48}+\sqrt{27}$
b)
$(3-\sqrt{2})^2$
c)
$\dfrac{4}{3-\sqrt{5}}$

Stuck? Peek at a hint

Use the largest perfect-square factor: $48=16\times3$ . For (b), write $(3-\sqrt{2})(3-\sqrt{2})$ and expand all four products.

K4 Three Ways to Solve

a)
Solve $x^2-2x=15$ by factoring.
b)
Solve $x^2+6x-4=0$ by completing the square. Give exact answers.
c)
Solve $2x^2-4x-3=0$ with the quadratic formula. Give exact answers in simplest form.

Stuck? Peek at a hint

Move every term to one side so the equation equals $0$ before you factor.

K5 Families & Systems

a)
Determine the equation of the parabola graphed below, in factored form. Each grid square is 1 unit.
b)
Find the member of the family $y=a(x-1)^2+3$ that passes through $(3,-5)$ .
c)
Solve the system $y=x^2-3x+1$ and $y=x-2$ .

Stuck? Peek at a hint

Write the family first, with one unknown $a$ , then substitute a known point to find $a$ .

Real World (A)

Optimization · Inverses in Context · Line Meets Parabola

A1 Muffin Pricing

A café in Kingston tracks its daily muffin sales. Each $0.25 increase in price means 8 fewer muffins sold.

$\text{Price (dollars)}$	$2.50$	$2.75$	$3.00$	$3.25$
$\text{Muffins sold}$	$160$	$152$	$144$	$136$

a)
Let $x$ be the number of $0.25 increases. Write the daily revenue $R(x)$ , then expand it to standard form.
b)
Determine the price that maximizes daily revenue, and the maximum revenue.
c)
The owner wants $500 a day from muffins. How many $0.25 increases will do it?

Stuck? Peek at a hint

Revenue is price times number sold. Write each one in terms of $x$ first.

A2 Skid Marks

On dry pavement, a car travelling at $v$ km/h needs about $d(v)=0.005v^2$ metres to brake to a stop, for $v\ge0$ .

a)
Determine $d^{-1}(x)$ . What do its input and its output represent?
b)
After a collision in a 70 km/h zone, police measure skid marks 32 m long. Use $d^{-1}$ to decide whether the driver was speeding.
c)
The equation $0.005v^2=32$ has two solutions. State both, and say which one the police can use and why.

Stuck? Peek at a hint

Write $x=0.005v^2$ and solve for $v$ .

A3 Uphill Golf

A golf ball is hit from the bottom of a hill, at the origin. Its path is $y=-0.02x^2+1.2x$ and the hillside is the line $y=0.2x$ , where $x$ is the horizontal distance and $y$ the height, both in metres. The sketch is not to scale.

a)
Determine the point where the ball lands on the hillside.
b)
Determine the greatest vertical distance between the ball and the hillside, and the value of $x$ where it occurs.
c)
A tree 10 m tall grows on the hillside at $x=40$ . Does the ball clear it?

Stuck? Peek at a hint

The ball meets the hill where $-0.02x^2+1.2x=0.2x$ . The vertical gap is the difference of the two heights.

Think It Through (T)

Error Analysis · Counterexamples · Tangent Lines

T1 Spot the Flaw

A student finds the inverse of $f(x)=2(x-3)^2-8$ , restricted to $x\le3$ :
"Swap $x$ and $y$ : $x=2(y-3)^2-8$ , so $\frac{x+8}{2}=(y-3)^2$ . Therefore $f^{-1}(x)=3+\sqrt{\frac{x+8}{2}}$ , with domain $x\le3$ and range $y\ge-8$ ."

a)
Evaluate $f(1)$ , then substitute the result into the student's formula. What does this show?
b)
Identify the two errors. Then write the correct $f^{-1}(x)$ with its domain and range.
c)
Evaluate $f^{-1}(10)$ and verify that $f(f^{-1}(10))=10$ .

Stuck? Peek at a hint

An inverse must send $f(1)$ back to $1$ .

T2 Line Meets Parabola

Let $p(x)=x^2-4x+5$ .

a)
A student claims: "The discriminant of $p$ is negative, so no line can meet the graph of $p$ twice." Give a counterexample and its two meeting points.
b)
Determine $k$ so that the line $y=2x+k$ is tangent to the graph of $p$ , and state the point of tangency.
c)
For which values of $k$ does the line $y=2x+k$ miss the graph of $p$ entirely?

Stuck? Peek at a hint

For (a), try a horizontal line. For (b), set $p(x)=2x+k$ and collect every term on one side.

Boss Level

Optional challenge

Vertices on a Line

For each real number $k$ , the parabola $P_k$ has equation $y=x^2-2kx+k^2+k-6$ .

a)
Write $P_k$ in vertex form. Find the equation of the line that contains the vertex of every $P_k$ .
b)
Determine all values of $k$ for which $P_k$ has two $x$ -intercepts.
c)
Determine the value of $k$ for which the two $x$ -intercepts of $P_k$ are 4 units apart.

Stuck? Peek at a hint

In (a), $x^2-2kx+k^2$ is already a perfect square. In (c), solve $(x-k)^2=6-k$ for $x$ .

Answer key

Final answers only. Try first, then check.

Core Skills

a) Vertex $(-3,5)$ ; axis $x=-3$ ; opens down
b) $-13$
c) Zeros $1$ and $-5$ ; axis $x=-2$ ; vertex $(-2,-27)$

a) $y=(x-5)^2-7$ ; minimum $-7$ at $x=5$
b) $y=2(x+3)^2-11$ ; minimum $-11$ at $x=-3$
c) $y=-3(x-2)^2+7$ ; maximum $7$ at $x=2$

a) $7\sqrt{3}$
b) $11-6\sqrt{2}$
c) $3+\sqrt{5}$

a) $x=5$ or $x=-3$
b) $x=-3\pm\sqrt{13}$
c) $x=\dfrac{2\pm\sqrt{10}}{2}$

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

Real World

a) $R(x)=(2.50+0.25x)(160-8x)=-2x^2+20x+400$
b) $3.75 per muffin; $450 per day
c) None: $R(x)=500$ gives $x^2-10x+50=0$ , whose discriminant is $-100<0$ . The most this model allows is $450 a day.

a) $d^{-1}(x)=\sqrt{200x}$ ; input: a braking distance in m; output: the speed in km/h
b) $d^{-1}(32)=80$ km/h, so yes: 10 km/h over the limit
c) $v=80$ or $v=-80$ . Only $80$ km/h: a speed can't be negative, so $-80$ is outside the domain $v\ge0$ . That restriction is also what makes $d^{-1}$ a function.

a) $(50,10)$ : 50 m out and 10 m up
b) $12.5$ m, at $x=25$ m
c) No. At $x=40$ the ball is 16 m high, but the tree stands on the hill at 8 m, so its top is at 18 m: the ball hits the tree 2 m below the top.

Think It Through

a) $f(1)=0$ , and the student's formula gives $5$ , not $1$ : it does not undo $f$ .
b) Wrong root: the outputs of $f^{-1}$ must be $\le3$ , so $f^{-1}(x)=3-\sqrt{\frac{x+8}{2}}$ . Domain and range not swapped: $D=[-8,\infty)$ , $R=(-\infty,3]$ .
c) $f^{-1}(10)=0$ ; $f(0)=2(9)-8=10$

a) e.g. $y=5$ , which meets the graph at $(0,5)$ and $(4,5)$
b) $k=-4$ ; tangent at $(3,2)$
c) $k<-4$

Boss Level

Boss

a) $y=(x-k)^2+k-6$ ; the vertices $(k,k-6)$ lie on $y=x-6$
b) $k<6$
c) $k=2$ (intercepts $0$ and $4$ )

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Quadratic Functions Cheat Sheet

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

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Most testedWrite $y=-2x^2+8x+3$ in vertex form and state the maximum value. (card 2)

1Three forms

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

$y=a(x-r)(x-s)$ : zeros $r,s$

axis of symmetry $x=(r+s)/2$

$y=ax^2+bx+c$ : $y$ -intercept $c$

Use $a>0$ opens up; $|a|>1$ is narrower than $y=x^2$ .

Trap $y=(x-2)(x+6)$ has zeros $2$ and $-6$ , so the axis is $x=-2$ , not $x=2$ .

Trap $3(x-2)^2\ne(3x-6)^2$ : square first, then multiply by $3$ .

2Completing the square

UseFactor $a$ from the $x^2$ and $x$ terms. Add and subtract $(b/2)^2$ inside. The subtracted number is multiplied by $a$ as it comes out.

Example $-2x^2+8x+3=-2(x^2-4x+4-4)+3=-2(x-2)^2+11$ : max $11$ at $x=2$ .

Trap $-2(x^2-4x+4-4)$ : the $-4$ comes out as $(-2)(-4)=+8$ , not $-4$ .

3Optimization

UseWrite the quantity to optimize as a quadratic, find the vertex, then answer the question that was asked.

Example $R=(20+2x)(100-5x)$ peaks at $x=5$ : price $20+2(5)=$ $30, revenue $2250.

Trap $x=5$ is the number of increases, not the price: substitute it back.

4Radicals

$\sqrt{ab}=\sqrt a\,\sqrt b\qquad(a-\sqrt b)(a+\sqrt b)=a^2-b$

UseSimplify with the largest perfect-square factor. Add like radicals only. Rationalize with the conjugate.

Example $\sqrt{48}=4\sqrt3$ ; $\sqrt{12}+\sqrt{27}=2\sqrt3+3\sqrt3=5\sqrt3$ .

Trap $(3+\sqrt2)^2=9+6\sqrt2+2=11+6\sqrt2$ , not $11$ .

Trap $(5-\sqrt2)(5+\sqrt2)=25-2=23$ , not $27$ .

5Inverse of a quadratic

$f^{-1}(x)=h\pm\sqrt{\dfrac{x-k}{a}}$

UseRestrict $f$ to $x\ge h$ (keep $+$ ) or $x\le h$ (keep $-$ ) so $f^{-1}$ is a function. The graph reflects over $y=x$ .

Example $f(x)=2(x-1)^2+3$ , $x\ge1$ : $f^{-1}(x)=1+\sqrt{(x-3)/2}$ .

TrapRestrict the domain with $h$ : vertex $(5,-3)$ needs $x\ge5$ , not $x\ge-3$ .

6Solving

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

UseGet $0$ on one side first. Factor if you can; otherwise complete the square or use the formula.

Trap $(x-2)(x+1)=4$ does not give $x-2=4$ : expand to $x^2-x-6=0$ , so $x=3$ or $-2$ .

TrapWith $b=-5$ , $b^2=(-5)^2=25$ , never $-25$ .

7The discriminant

$D=b^2-4ac$ :

$D>0$	two real roots
$D=0$	one (double) root
$D<0$	no real roots

TipIn context, reject roots that make no sense, like a negative time or length.

8Families

UseSame vertex: $y=a(x-h)^2+k$ . Same zeros: $y=a(x-r)(x-s)$ . Substitute one more point to find $a$ .

ExampleZeros $-3$ and $2$ , through $(0,12)$ : $12=a(3)(-2)$ , so $y=-2(x+3)(x-2)$ .

TrapA zero of $-3$ gives the factor $(x+3)$ , not $(x-3)$ .

9Line and parabola

UseSet the equations equal and collect every term on one side. The discriminant of that equation counts the intersections: $2$ secant, $1$ tangent, $0$ none.

Example $x^2+x-1=3x+2$ : $x^2-2x-3=0$ , so $(3,11)$ and $(-1,-1)$ .

TrapUse the discriminant of the combined equation, not of the parabola alone.

Which method?

It factors with small integers Factor

Need the vertex Complete the square

It doesn't factor nicely Quadratic formula

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