Maximum and Minimum of a Quadratic

The vertex of a parabola is its highest or lowest point. That makes quadratics perfect for optimization: finding the greatest height, the largest area, or the most revenue. This page shows two ways to find the vertex and how to use it in word problems.

Key ideas

Maximum or minimum?

For $f(x) = ax^2 + bx + c$:

• $a \gt 0$: the parabola opens up, so the vertex is a minimum.
• $a \lt 0$: it opens down, so the vertex is a maximum.

The $y$-coordinate of the vertex is the max or min value. The $x$-coordinate tells you where (or when) it happens.

Method 1: Complete the square

Completing the square gives vertex form $a(x - h)^2 + k$, with vertex $(h, k)$. This always works.

Method 2: Average the zeros

A parabola is symmetric about its axis of symmetry, the vertical line through the vertex. So the vertex is exactly halfway between the zeros:

$$x_{\text{vertex}} = \frac{r + s}{2}$$

Then substitute that $x$ into the function to get the max or min value. This method is quick when the quadratic factors, but it needs the function to have zeros.

The vertex sits on the axis of symmetry, halfway between the zeros.

A shortcut that comes from completing the square: the vertex is at $x = -\dfrac{b}{2a}$.

Solving word problems

1. Write the quadratic (if it isn’t given), with clear units.
2. Find the vertex.
3. Answer the question asked, in context: is it asking for the max value, or when it happens?

Worked examples

Example 1: Averaging the zeros

Find the maximum value of $f(x) = -2x^2 + 12x - 10$.

Solution. Factor to find the zeros:

$$f(x) = -2(x^2 - 6x + 5) = -2(x - 1)(x - 5)$$

The zeros are $1$ and $5$, so the vertex is at $x = \dfrac{1 + 5}{2} = 3$.

$$f(3) = -2(9) + 36 - 10 = 8$$

Since $a \lt 0$, the maximum value is $8$, at $x = 3$.

Example 2: When there are no zeros

Find the minimum value of $g(x) = x^2 - 4x + 7$.

Solution. The discriminant is $(-4)^2 - 4(1)(7) = -12$, so there are no zeros to average. Complete the square instead:

$$g(x) = x^2 - 4x + 4 - 4 + 7 = (x - 2)^2 + 3$$

The minimum value is $3$, at $x = 2$.

Example 3: A projectile

A ball is kicked from a height of $2$ m. Its height in metres after $t$ seconds is $h(t) = -5t^2 + 30t + 2$. Find the maximum height and when it happens.

Solution. Use $t = -\dfrac{b}{2a}$:

$$t = -\frac{30}{2(-5)} = 3$$ $$h(3) = -5(9) + 30(3) + 2 = -45 + 90 + 2 = 47$$

The ball reaches a maximum height of $47$ m after $3$ seconds.

Example 4: Maximizing revenue

A bike shop rents $120$ bikes a day at $20 each. For every $1 increase in price, $4$ fewer bikes are rented. What price gives the most revenue?

Solution. Let $n$ be the number of $1 increases. Then the price is $20 + n$ dollars and the number rented is $120 - 4n$:

$$R(n) = (20 + n)(120 - 4n)$$

This is already in factored form. The zeros are $n = -20$ (from $20 + n = 0$) and $n = 30$ (from $120 - 4n = 0$). The maximum is halfway between:

$$n = \frac{-20 + 30}{2} = 5$$

So the best price is $25 (that’s $20 + 5$), with $120 - 4(5) = 100$ bikes rented. The revenue is $25 \times 100 = 2500$, or $2500.

Common mistakes

Giving $x$ when the question asks for the value. In Example 3, “the maximum height” is $47$ m, not $3$.

Averaging zeros that don’t exist. If the discriminant is negative, there are no zeros. Complete the square or use $x = -\tfrac{b}{2a}$.

Getting the sign of $-\tfrac{b}{2a}$ wrong. For $h(t) = -5t^2 + 30t + 2$, $-\tfrac{30}{2(-5)} = 3$, which is positive.

Mixing up maximum and minimum. Check the sign of $a$: negative means a maximum.

Ignoring the context. In Example 4, the answer is a price ($25), not the value of $n$.

Practice

1. (Warm-up) Does $f(x) = 3(x - 4)^2 + 1$ have a maximum or a minimum? What is it, and where does it happen?

Solution

$a = 3 \gt 0$, so a minimum. The minimum value is $1$, at $x = 4$.

2. (Warm-up) A parabola has zeros at $-2$ and $6$. What is the equation of its axis of symmetry?

Solution

$x = \dfrac{-2 + 6}{2} = 2$, so the axis is $x = 2$.

3. (Warm-up) Find the maximum value of $f(x) = -(x - 1)(x - 7)$.

Solution

The zeros are $1$ and $7$, so the vertex is at $x = 4$:

$$f(4) = -(3)(-3) = 9$$

The maximum value is $9$.

4. (Core) Find the minimum value of $f(x) = 2x^2 - 8x + 3$.

Solution

$x = -\dfrac{-8}{2(2)} = 2$, and $f(2) = 8 - 16 + 3 = -5$.

The minimum value is $-5$, at $x = 2$.

5. (Core) Find the maximum value of $f(x) = -x^2 + 6x - 2$ by completing the square.

Solution

$$\begin{aligned} f(x) &= -(x^2 - 6x) - 2 \\ &= -(x^2 - 6x + 9 - 9) - 2 \\ &= -(x - 3)^2 + 9 - 2 \\ &= -(x - 3)^2 + 7 \end{aligned}$$

The maximum value is $7$, at $x = 3$.

6. (Core) A farmer has $60$ m of fencing to make a rectangular pen against a long barn wall, so only three sides need fencing. What dimensions give the largest area?

Solution

Let $w$ be the width of each side touching the wall. The side opposite the wall is $60 - 2w$.

$$A(w) = w(60 - 2w)$$

The zeros are $w = 0$ and $w = 30$, so the maximum is at $w = 15$.

The pen should be $15$ m by $30$ m (the $30$ m side opposite the wall), for an area of $450$ m².

7. (Core) A ball’s height in metres after $t$ seconds is $h(t) = -4.9t^2 + 24.5t + 1$. Find its maximum height, to one decimal place, and when it happens.

Solution

$$t = -\frac{24.5}{2(-4.9)} = 2.5$$$$h(2.5) = -4.9(6.25) + 24.5(2.5) + 1 = -30.625 + 61.25 + 1 = 31.625$$

The maximum height is about $31.6$ m, after $2.5$ seconds.

8. (Challenge) Two numbers add up to $20$. What is the largest possible value of their product?

Solution

If one number is $x$, the other is $20 - x$, and the product is $P(x) = x(20 - x)$.

The zeros are $0$ and $20$, so the maximum is at $x = 10$: $P(10) = 10 \times 10 = 100$.

9. (Challenge) A theatre sells $300$ tickets at $15 each. Each $1 increase in price means $12$ fewer tickets sold. What price maximizes revenue, and what is the maximum revenue?

Solution

With $n$ increases of $1, $R(n) = (15 + n)(300 - 12n)$.

The zeros are $n = -15$ and $n = 25$, so the maximum is at $n = \dfrac{-15 + 25}{2} = 5$.

The best price is $20, with $300 - 12(5) = 240$ tickets sold. The maximum revenue is $20 \times 240 = 4800$, or $4800.