2025-02-04

Pages 22-24 (#1-17)

1

Evaluate, where $f(x)=2-3x$
a) $f(2)=2-3(2)=-4$
b) $f(0)=2-3(0)=2$
c) $f(-4)=2-3(-4)=14$
d) $f(0.5)=2-3(0.5)=0.5$
e) $f(a)=2-3a$
f) $f(3b)=2-3(3b)=2-9b$

2

a) 2
b) 4
c) -5
d) -3 or -4

3

Milk is leaking from a carton at a rate of 3 mL/min. There is 1.2 L of milk in the carton at 11:00 a.m.

a) Function notation:

$$L(t)=1200-3t$$

b) At 1:00 p.m. ($t=120$ min):

$$L(120)=1200-3(120)=840\text{ mL}$$

c) Time when 450 mL is left:

$$\begin{array}{rl}& L(t)=450\\ & 1200-3t=450\\ & -3t=-750\\ & t=250\\ & \therefore \text{At 3:10 p.m., 450 mL remains}\end{array}$$

4

Solve $h(x)=2x^2-12x+18$ a) Vertex form: \begin{align} h(x) &= 2(x^2 - 6x) + 18 \\ &= 2(x^2 - 6x + 9 - 9) + 18 \\ &= 2((x - 3)^2 - 9) + 18 \\ &= 2(x - 3)^2 - 18 + 18 \\ &= 2(x - 3)^2 \end{align} Vertex: $(3,0)$ b) Solve $h(x)=0$: $2x^2-12x+18=0$ Factor: $2(x^2-6x+9)=0$ $2(x-3{)}^2=0$ $x=3$

5

Find inverse of $g(x)=4x-7$: \begin{align} y &= 4x - 7 \\ x &= 4y - 7 \\ x + 7 &= 4y \\ y &= \frac{x+7}{4} \\ \therefore g^{-1}(x) = \frac{x+7}{4} \end{align} # 6 Solve $f(x)=-x^2+10x-16$ a) Vertex form: \begin{align} f(x) &= -(x^2 - 10x) - 16 \\ &= -(x^2 - 10x + 25 - 25) - 16 \\ &= -((x - 5)^2 - 25) - 16 \\ &= -(x - 5)^2 + 25 - 16 \\ &= -(x - 5)^2 + 9 \end{align} Vertex: $(5,9)$ b) Solve $f(x)=0$: $-x^2+10x-16=0$ Quadratic formula: $x=\frac{-10\pm \sqrt{(10{)}^2-4(-1)(-16)}}{2(-1)}$ $x=\frac{-10\pm \sqrt{100-64}}{-2}$ $x=\frac{-10\pm \sqrt{36}}{-2}$ $x=\frac{-10\pm 6}{-2}$ $x=8,2$

7

Find inverse of $h(x)=5x+2$: \begin{align} y &= 5x + 2 \\ x &= 5y + 2 \\ x - 2 &= 5y \\ y &= \frac{x - 2}{5} \\ \therefore h^{-1}(x) = \frac{x - 2}{5} \end{align}

8

Solve $g(x)=-3x^2+18x-27$ a) Vertex form: \begin{align} g(x) &= -3(x^2 - 6x) - 27 \\ &= -3(x^2 - 6x + 9 - 9) - 27 \\ &= -3((x - 3)^2 - 9) - 27 \\ &= -3(x - 3)^2 + 27 - 27 \\ &= -3(x - 3)^2 \end{align} Vertex: $(3,0)$ b) Solve $g(x)=0$: $-3(x-3{)}^2=0$ $x-3=0$ $x=3$

9

Find inverse of $f(x)=2x-5$: \begin{align} y &= 2x - 5 \\ x &= 2y - 5 \\ x + 5 &= 2y \\ y &= \frac{x+5}{2} \\ \therefore f^{-1}(x) = \frac{x+5}{2} \end{align}

10

Solve $h(x)=x^2-4x+3$ a) Vertex form: \begin{align} h(x) &= (x^2 - 4x) + 3 \\ &= (x^2 - 4x + 4 - 4) + 3 \\ &= ((x - 2)^2 - 4) + 3 \\ &= (x - 2)^2 - 1 \end{align} Vertex: $(2,-1)$ b) Solve $h(x)=0$: $(x-2{)}^2-1=0$ $(x-2{)}^2=1$ $x-2=\pm 1$ $x=3,1$

11

Find inverse of $g(x)=3x+4$: \begin{align} y &= 3x + 4 \\ x &= 3y + 4 \\ x - 4 &= 3y \\ y &= \frac{x - 4}{3} \\ \therefore g^{-1}(x) = \frac{x - 4}{3} \end{align}

12

a)

$$f(k)=50+0.15k$$

b)

$$f(480)=50+0.15(480)=120.8$$

c) Solving $f(k)=80$:

$$\begin{array}{rl}& 50+0.15k=80\\ & 0.15k=30\\ & k=200\end{array}$$

13

a)

$$f(x)=x(24-3x)$$

b) $f(3)=45,f(-5)=-195,f(10)=-60$

c) Finding max value:

$$\begin{array}{rl}& f(x)=x(24-3x)\\ & =-3x^2+24x\\ & -3(x^2-8x)=0\\ & -3(x^2-8x+16-16)=48\\ & -3(x-4{)}^2=48\\ & v(4,48)\\ & \therefore \text{Maximum value is 48 at }x=4\end{array}$$

14

Solve for $f(x)=2x^2-8x+5$

a) Vertex form:

$$\begin{array}{rl}f(x)& =2(x^2-4x)+5\\ & =2(x^2-4x+4-4)+5\\ & =2((x-2{)}^2-4)+5\\ & =2(x-2{)}^2-8+5\\ & =2(x-2{)}^2-3\end{array}$$

Vertex: $(2,-3)$

b) Solve $f(x)=0$:

$$\begin{array}{rl}& 2x^2-8x+5=0\\ & \text{Use quadratic formula: }x=\frac{-(-8)\pm \sqrt{(-8{)}^2-4(2)(5)}}{2(2)}\\ x& =\frac{8\pm \sqrt{64-40}}{4}\\ x& =\frac{8\pm \sqrt{24}}{4}\\ x& =\frac{8\pm 2\sqrt{6}}{4}\\ x& =2\pm \frac{\sqrt{6}}{2}\end{array}$$

15

Find inverse of $f(x)=3x-7$:

$$\begin{array}{rl}y& =3x-7\\ x& =3y-7\\ x+7& =3y\\ y& =\frac{x+7}{3}\\ \therefore f^{-1}(x)=\frac{x+7}{3}& \end{array}$$

16

Solve $g(x)=5x^2-20x+15$

a) Vertex form:

$$\begin{array}{rl}g(x)& =5(x^2-4x)+15\\ & =5(x^2-4x+4-4)+15\\ & =5((x-2{)}^2-4)+15\\ & =5(x-2{)}^2-20+15\\ & =5(x-2{)}^2-5\end{array}$$

Vertex: $(2,-5)$

b) Solve $g(x)=0$:

$$\begin{array}{rl}& 5x^2-20x+15=0\\ & \text{Quadratic formula: }x=\frac{20\pm \sqrt{(-20{)}^2-4(5)(15)}}{2(5)}\\ x& =\frac{20\pm \sqrt{400-300}}{10}\\ x& =\frac{20\pm \sqrt{100}}{10}\\ x& =\frac{20\pm 10}{10}\\ x& =3,1\end{array}$$

17

Solve $h(x)=-2x^2+12x-10$

a) Vertex form:

$$\begin{array}{rl}h(x)& =-2(x^2-6x)-10\\ & =-2(x^2-6x+9-9)-10\\ & =-2((x-3{)}^2-9)-10\\ & =-2(x-3{)}^2+18-10\\ & =-2(x-3{)}^2+8\end{array}$$

Vertex: $(3,8)$

b) Solve $h(x)=0$:

$$\begin{array}{rl}& -2x^2+12x-10=0\\ & \text{Quadratic formula: }x=\frac{-12\pm \sqrt{(12{)}^2-4(-2)(-10)}}{2(-2)}\\ x& =\frac{-12\pm \sqrt{144-80}}{-4}\\ x& =\frac{-12\pm \sqrt{64}}{-4}\\ x& =\frac{-12\pm 8}{-4}\\ x& =5,1\end{array}$$