Quadratic Functions and Equations: practice solutions, Grade 11

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Practice solutions Grade 11 : Quadratic Functions and Equations — Zyro the alien explorer of Planète Maths

Written solutions to the chapter problems. Check each step, then correct yourself.

Sommaire

1 Reading the coefficients ★★★

(a) $a = -3$, $b = 12$, $c = -2$. Since $a < 0$, the parabola opens downward.

(b) The axis is $x = -\dfrac{12}{2 \cdot (-3)} = 2$. Then $f(2) = -12 + 24 - 2 = 10$, so the vertex is $(2, 10)$, a maximum.

2 Reading the vertex form ★★★

Compare with $a(x - h)^2 + k$: $a = 2$, $h = -4$, $k = -9$. The vertex is $(-4, -9)$ and the axis is $x = -4$.

Since $a = 2 > 0$, the parabola opens upward and has a minimum value of $-9$.

The y-intercept is $g(0) = 2 \cdot 16 - 9 = 23$.

3 Factoring simple trinomials ★★★

(a) Product $20$, sum $9$: the numbers are $4$ and $5$, so $x^2 + 9x + 20 = (x + 4)(x + 5)$.

(b) Product $-12$, sum $-1$: the numbers are $-4$ and $3$, so $(x - 4)(x + 3)$.

Solving $(x + 4)(x + 5) = 0$ gives $x = -4$ or $x = -5$.

4 Common factor ★★★

Factor out $5x$: $5x(x - 7) = 0$.

By the zero-product property, $5x = 0$ or $x - 7 = 0$, so $x = 0$ or $x = 7$.

Check: $5 \cdot 49 - 35 \cdot 7 = 245 - 245 = 0$.

5 Square root method ★★★

Divide by $3$: $(x - 2)^2 = 16$.

Take square roots: $x - 2 = 4$ or $x - 2 = -4$, so $x = 6$ or $x = -2$.

6 Using the discriminant ★★★

(a) $\Delta = 9 - 40 = -31 < 0$: no real solution.

(b) $\Delta = 36 - 36 = 0$: one real solution (a double root, $x = -3$).

7 Reading a graph for an inequality ★★★

(a) At $x = -1$ and $x = 4$, the zeros of the two factors.

(b) The graph is below the x-axis between the roots: $-1 < x < 4$.

8 Completing the square to get the vertex ★★★

Half of $8$ is $4$, and $4^2 = 16$. So $y = (x^2 + 8x + 16) - 16 + 3 = (x + 4)^2 - 13$.

The vertex is $(-4, -13)$.

9 Vertex form with a negative leading coefficient ★★★

Factor $-2$ from the first two terms: $y = -2(x^2 - 6x) - 7$.

Complete the square inside: $x^2 - 6x = (x - 3)^2 - 9$. So $y = -2\big((x - 3)^2 - 9\big) - 7 = -2(x - 3)^2 + 18 - 7 = -2(x - 3)^2 + 11$.

The vertex is $(3, 11)$, and since $a < 0$ the maximum value is $11$.

10 Factoring by grouping ★★★

We need two numbers with product $2 \cdot (-12) = -24$ and sum $5$: $8$ and $-3$.

$2x^2 + 8x - 3x - 12 = 2x(x + 4) - 3(x + 4) = (2x - 3)(x + 4)$.

So $x = \dfrac{3}{2}$ or $x = -4$.

11 Quadratic formula with a radical ★★★

$a = 1$, $b = -4$, $c = -3$, so $\Delta = 16 + 12 = 28$.

$x = \dfrac{4 \pm \sqrt{28}}{2} = \dfrac{4 \pm 2\sqrt{7}}{2} = 2 \pm \sqrt{7}$.

Numerically $x \approx 4.65$ or $x \approx -0.65$.

12 A double root ★★★

One real solution means $\Delta = 0$: $k^2 - 4 \cdot 25 = 0$, so $k^2 = 100$ and $k = 10$ or $k = -10$.

If $k = 10$: $x^2 + 10x + 25 = (x + 5)^2 = 0$, so $x = -5$.

If $k = -10$: $(x - 5)^2 = 0$, so $x = 5$.

13 Building a function from its roots ★★★

Use $y = a(x - 2)(x - 7)$. At $x = 0$: $a \cdot (-2)(-7) = 14a = 28$, so $a = 2$.

$y = 2(x^2 - 9x + 14) = 2x^2 - 18x + 28$.

Check with $x = 3$: $2 \cdot 1 \cdot (-4) = -8$ and $18 - 54 + 28 = -8$. Both agree.

14 Two quadratic inequalities ★★★

(a) $x^2 - 5x + 4 = (x - 1)(x - 4)$. The parabola opens upward and is positive outside its roots: $x < 1$ or $x > 4$, that is $(-\infty, 1) \cup (4, \infty)$.

(b) $2x^2 + x - 6 = (2x - 3)(x + 2)$, with roots $\dfrac{3}{2}$ and $-2$. The parabola is negative or zero between them: $-2 \leq x \leq \dfrac{3}{2}$, that is $\left[-2, \dfrac{3}{2}\right]$.

15 The patio ★★★

Let $w$ be the width in feet. Then $w(w + 4) = 252$, so $w^2 + 4w - 252 = 0$.

$\Delta = 16 + 1008 = 1024 = 32^2$, so $w = \dfrac{-4 \pm 32}{2}$, that is $w = 14$ or $w = -18$. A width cannot be negative, so $w = 14$.

The patio is $14$ ft wide and $18$ ft long (about $4.3$ m by $5.5$ m). Check: $14 \cdot 18 = 252$.

16 Complex solutions ★★★

$\Delta = 36 - 52 = -16$, so $\sqrt{\Delta} = 4i$. Then $x = \dfrac{6 \pm 4i}{2} = 3 \pm 2i$.

Check $3 + 2i$: $(3 + 2i)^2 = 9 + 12i - 4 = 5 + 12i$; $-6(3 + 2i) = -18 - 12i$; the sum $5 + 12i - 18 - 12i + 13 = 0$.

17 A rock thrown from a cliff ★★★

(a) Solve $-16t^2 + 24t + 40 = 0$. Dividing by $-8$: $2t^2 - 3t - 5 = 0$, which factors as $(2t - 5)(t + 1) = 0$. So $t = 2.5$ or $t = -1$. Time cannot be negative: the rock hits the ground after 2.5 seconds.

(b) The vertex is at $t = -\dfrac{24}{2 \cdot (-16)} = 0.75$ s, where $h(0.75) = -9 + 18 + 40 = 49$. The maximum height is $49$ feet (about $14.9$ m), reached after $0.75$ s.

18 Revenue and price ★★★

(a) $R(p) = -4p^2 + 120p$. The vertex is at $p = -\dfrac{120}{2 \cdot (-4)} = 15$, and $R(15) = 15 \cdot 60 = 900$. The best price is \$15, for a revenue of \$900.

(b) Solve $-4p^2 + 120p \geq 800$, so $-4p^2 + 120p - 800 \geq 0$. Divide by $-4$ and reverse the inequality: $p^2 - 30p + 200 \leq 0$, that is $(p - 10)(p - 20) \leq 0$. So $10 \leq p \leq 20$.

The revenue is at least \$800 for prices between \$10 and \$20.

19 A line and a parabola ★★★

Set the expressions equal: $x^2 - 2x - 3 = x + 1$, so $x^2 - 3x - 4 = 0$, that is $(x - 4)(x + 1) = 0$.

$x = 4$ gives $y = 5$; $x = -1$ gives $y = 0$.

The intersection points are $(4, 5)$ and $(-1, 0)$. Check on the parabola: $16 - 8 - 3 = 5$ and $1 + 2 - 3 = 0$.

20 Roots that differ by 6 ★★★

Call the roots $r$ and $r + 6$. Their sum is $10$ (the opposite of the coefficient of $x$), so $2r + 6 = 10$ and $r = 2$.

The roots are $2$ and $8$, and $c = 2 \cdot 8 = 16$.

Check with the discriminant: $100 - 4 \cdot 16 = 36$, and $\dfrac{10 \pm 6}{2}$ gives $8$ and $2$.

21 Always positive ★★★

The parabola opens upward, so the expression is always positive exactly when the graph stays above the x-axis, that is, when there is no real root: $\Delta < 0$.

$\Delta = 16 - 4k < 0$ gives $k > 4$.

With the vertex: $x^2 - 4x + k = (x - 2)^2 + k - 4$. The minimum value is $k - 4$, which is positive when $k > 4$. Both methods agree.