State the vertex, the axis of symmetry and the maximum or minimum value of $y = -3(x - 4)^2 + 7$.

Vertex $(4, 7)$, axis $x = 4$, maximum value $7$.

Write $y = x^2 + 8x + 5$ in vertex form.

$y = (x + 4)^2 - 11$: vertex $(-4, -11)$, minimum $-11$.

Write $y = -2x^2 + 12x - 5$ in vertex form and state its maximum value.

$y = -2(x - 3)^2 + 13$; maximum $13$ at $x = 3$.

Write $y = 3x^2 + 9x + 2$ in vertex form.

$y = 3\left(x + \frac{3}{2}\right)^2 - \frac{19}{4}$: vertex $\left(-\frac{3}{2}, -\frac{19}{4}\right)$, minimum $-\frac{19}{4}$.

A theatre sells $400$ tickets a night at $12$ dollars each. For every $1$ dollar increase in price, it sells $20$ fewer tickets. What price gives the maximum revenue, and what is that revenue?

A price of $16$ dollars gives the maximum revenue, $5120$ dollars.

Grade 11 · MCR3U · Functions

MCR3U: Max/Min and Vertex Form

Vertex form, $y = a(x - h)^2 + k$ , puts the most useful facts about a parabola in plain sight. The vertex is $(h, k)$ , the axis of symmetry is $x = h$ , and $k$ is the maximum or minimum value. Getting there from standard form takes one technique: completing the square.

How It Works

Reading vertex form. If $a > 0$ the parabola opens up and $k$ is a minimum; if $a < 0$ it opens down and $k$ is a maximum. Watch the sign inside the bracket: $y = 2(x + 3)^2 - 5$ has $h = -3$ , so the vertex is $(-3, -5)$ .

Completing the square turns $y = ax^2 + bx + c$ into vertex form. Factor $a$ out of the first two terms. Inside the bracket, take half the coefficient of $x$ , square it, then add and subtract that number. The first three terms in the bracket are now a perfect square.

Bring the extra term out. The number you subtracted is still multiplied by $a$ , so multiply it by $a$ as it leaves the bracket. Forgetting that is where most errors happen.

For example, $y = 2x^2 - 12x + 7 = 2(x^2 - 6x + 9 - 9) + 7 = 2(x - 3)^2 - 18 + 7 = 2(x - 3)^2 - 11$ . The minimum is $-11$ , at $x = 3$ .

A check on the vertex. The axis of symmetry is $x = -\frac{b}{2a}$ , and substituting that value gives $k$ . Use it to check a completed square, or when a question asks only for the maximum or minimum. If the quadratic factors, the vertex is also halfway between the zeros.

Max and min word problems. Write the quantity as a quadratic in one variable, then find its vertex. The $x$ -coordinate tells you where the best value happens, and $k$ tells you what it is. Check the answer against the context: a price or a length cannot be negative.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
State the vertex, the axis of symmetry and the maximum or minimum value of $y = -3(x - 4)^2 + 7$ .
Show solution
1. $h = 4$ and $k = 7$ , so the vertex is $(4, 7)$ and the axis is $x = 4$ .
2. $a = -3 < 0$ , so the parabola opens down and $7$ is a maximum.
Answer: Vertex $(4, 7)$ , axis $x = 4$ , maximum value $7$ .
Easy
Write $y = x^2 + 8x + 5$ in vertex form.
Show solution
1. Half of $8$ is $4$ , and $4^2 = 16$ . Add and subtract it: $y = (x^2 + 8x + 16) - 16 + 5$ .
2. $y = (x + 4)^2 - 11$ .
Answer: $y = (x + 4)^2 - 11$ : vertex $(-4, -11)$ , minimum $-11$ .
Medium
Write $y = -2x^2 + 12x - 5$ in vertex form and state its maximum value.
Show solution
1. Factor $-2$ from the first two terms: $y = -2(x^2 - 6x) - 5$ .
2. Half of $-6$ is $-3$ , and $(-3)^2 = 9$ : $y = -2(x^2 - 6x + 9 - 9) - 5$ .
3. Bring $-9$ out, multiplied by $-2$ : $y = -2(x - 3)^2 + 18 - 5 = -2(x - 3)^2 + 13$ .
4. Check: $x = -\frac{12}{2(-2)} = 3$ , and $-2(9) + 36 - 5 = 13$ .
Answer: $y = -2(x - 3)^2 + 13$ ; maximum $13$ at $x = 3$ .
Medium
Write $y = 3x^2 + 9x + 2$ in vertex form.
Show solution
1. Factor $3$ from the first two terms: $y = 3(x^2 + 3x) + 2$ .
2. Half of $3$ is $\frac{3}{2}$ , and its square is $\frac{9}{4}$ : $y = 3\left(x^2 + 3x + \frac{9}{4} - \frac{9}{4}\right) + 2$ .
3. Bring $-\frac{9}{4}$ out, multiplied by $3$ : $y = 3\left(x + \frac{3}{2}\right)^2 - \frac{27}{4} + 2 = 3\left(x + \frac{3}{2}\right)^2 - \frac{19}{4}$ .
Answer: $y = 3\left(x + \frac{3}{2}\right)^2 - \frac{19}{4}$ : vertex $\left(-\frac{3}{2}, -\frac{19}{4}\right)$ , minimum $-\frac{19}{4}$ .
Test level
A theatre sells $400$ tickets a night at $12$ dollars each. For every $1$ dollar increase in price, it sells $20$ fewer tickets. What price gives the maximum revenue, and what is that revenue?
Show solution
1. Let $x$ be the number of $1$ dollar increases. Price: $12 + x$ . Tickets: $400 - 20x$ .
2. Revenue: $R = (12 + x)(400 - 20x) = -20x^2 + 160x + 4800$ .
3. Complete the square: $R = -20(x^2 - 8x + 16 - 16) + 4800 = -20(x - 4)^2 + 320 + 4800 = -20(x - 4)^2 + 5120$ .
4. The maximum is at $x = 4$ : a price of $16$ dollars, selling $400 - 80 = 320$ tickets.
Answer: A price of $16$ dollars gives the maximum revenue, $5120$ dollars.

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Martian Lab teaches this in the Maximum & Minimum of Quadratic Functions lesson of its MCR3U Quadratic Functions unit. Watch it, then take the Quadratic Functions practice tests to see if it stuck.

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More Quadratic Functions topics: Linear-Quadratic Systems · Quadratic Word Problems · Simplifying Radicals

More MCR3U topics: Inverse Functions · Transformations of Functions · Sine Law and the Ambiguous Case · Domain, Range and Function Notation · Simplifying Rational Expressions · Operations with Rational Expressions · Exponent Laws and Rational Exponents · Graphing Exponential Functions · Exponential Growth and Decay · Special Angles and the CAST Rule · Cosine Law and 3D Problems · Transformations of Sine and Cosine · Sinusoidal Word Problems · Arithmetic Sequences and Series · Geometric Sequences and Series · Pascal's Triangle and the Binomial Theorem · Simple and Compound Interest · Annuities: Future and Present Value

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