QUESTION IMAGE
Question
2.9 rate of change with equations - day 4
find the average rate of change for each equation:
- ( f(x) = x^2 - 3x + 12 ) from ( f(1) ) to ( f(2) )
- ( y = x + 8 ) on (0, 6)
- ( g(x) = -2x^2 - 2x + 18 ) from ( g(3) ) to ( g(4) )
- ( y = x - 1 ) on (-4, 8)
Step1: Define average rate of change formula
The average rate of change of a function $h(x)$ from $x=a$ to $x=b$ is $\frac{h(b)-h(a)}{b-a}$.
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For Problem 1: $f(x)=x^2-3x+12$, from $f(1)$ to $f(2)$
Step1: Calculate $f(1)$
$f(1)=(1)^2-3(1)+12=1-3+12=10$
Step2: Calculate $f(2)$
$f(2)=(2)^2-3(2)+12=4-6+12=10$
Step3: Compute average rate
$\frac{f(2)-f(1)}{2-1}=\frac{10-10}{1}=0$
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For Problem 2: $y=x+8$, over $[0,6]$
Step1: Calculate $y(0)$
$y(0)=0+8=8$
Step2: Calculate $y(6)$
$y(6)=6+8=14$
Step3: Compute average rate
$\frac{y(6)-y(0)}{6-0}=\frac{14-8}{6}=\frac{6}{6}=1$
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For Problem 3: $g(x)=-2x^2-2x+18$, from $g(3)$ to $g(4)$
Step1: Calculate $g(3)$
$g(3)=-2(3)^2-2(3)+18=-18-6+18=-6$
Step2: Calculate $g(4)$
$g(4)=-2(4)^2-2(4)+18=-32-8+18=-22$
Step3: Compute average rate
$\frac{g(4)-g(3)}{4-3}=\frac{-22-(-6)}{1}=-16$
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For Problem 4: $y=x-1$, over $[-4,8]$
Step1: Calculate $y(-4)$
$y(-4)=-4-1=-5$
Step2: Calculate $y(8)$
$y(8)=8-1=7$
Step3: Compute average rate
$\frac{y(8)-y(-4)}{8-(-4)}=\frac{7-(-5)}{12}=\frac{12}{12}=1$
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- $0$
- $1$
- $-16$
- $1$