Question

given the graphs of y = f(x) and y = g(x) shown below and h(x)=\frac{f(x)}{g(x)}, determine the value of h(-8).

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $h(x)=\frac{f(x)}{g(x)}$, then $h^{\prime}(x)=\frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{[g(x)]^{2}}$.

Step2: Find $f(-8)$, $f^{\prime}(-8)$, $g(-8)$ and $g^{\prime}(-8)$ from the graphs

• From the graph of $y = f(x)$: When $x=-8$, $f(-8)= - 2$. The slope of $y = f(x)$ at $x = - 8$ (using the two - point formula for slope on the line segment of $y = f(x)$ containing $x=-8$) is $f^{\prime}(-8)=\frac{2+2}{-5 + 8}=\frac{4}{3}$.
• From the graph of $y = g(x)$: When $x=-8$, $g(-8)=3$. The slope of $y = g(x)$ at $x=-8$ (using the two - point formula for slope on the line segment of $y = g(x)$ containing $x=-8$) is $g^{\prime}(-8)=\frac{6 - 3}{-5+8}=1$.

Step3: Substitute values into the quotient - rule formula

Substitute $x=-8$ into $h^{\prime}(x)=\frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{[g(x)]^{2}}$.
$h^{\prime}(-8)=\frac{f^{\prime}(-8)g(-8)-f(-8)g^{\prime}(-8)}{[g(-8)]^{2}}$.
$h^{\prime}(-8)=\frac{\frac{4}{3}\times3-(-2)\times1}{3^{2}}$.
$h^{\prime}(-8)=\frac{4 + 2}{9}=\frac{6}{9}=\frac{2}{3}$.

Answer:

$\frac{2}{3}$