Question

refer to the graph of y = f(x)=x² + x shown. a) find the slope of the secant line joining (-2, f(-2)) and (0, f(0)). b) find the slope of the secant line joining (-2, f(-2)) and (-2 + h, f(-2 + h)). c) find the slope of the graph at (-2, f(-2)). d) find the equation of the tangent line to the graph at (-2, f(-2)). a) the slope is - 1. (type an integer or a simplified fraction.) b) the slope is

Explanation:

Step1: Calculate $f(-2)$ and $f(0)$

Given $f(x)=x^{2}+x$, then $f(-2)=(-2)^{2}+(-2)=4 - 2=2$, $f(0)=0^{2}+0 = 0$.

Step2: Use slope - formula for secant line

The slope $m$ of the secant line joining $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2 - y_1}{x_2 - x_1}$. For $x_1=-2,y_1 = f(-2)=2,x_2 = 0,y_2=f(0)=0$, we have $m=\frac{0 - 2}{0-(-2)}=\frac{-2}{2}=-1$.

Step3: Calculate $f(-2 + h)$

$f(-2 + h)=(-2 + h)^{2}+(-2 + h)=4-4h+h^{2}-2 + h=h^{2}-3h + 2$.

Step4: Use slope - formula for secant line with $(-2,f(-2))$ and $(-2 + h,f(-2 + h))$

The slope of the secant line joining $(-2,f(-2))$ and $(-2 + h,f(-2 + h))$ is $m=\frac{f(-2 + h)-f(-2)}{(-2 + h)-(-2)}$. Since $f(-2)=2$ and $f(-2 + h)=h^{2}-3h + 2$, then $m=\frac{h^{2}-3h + 2-2}{h}=\frac{h^{2}-3h}{h}=h - 3$.

Step5: Find the slope of the graph at $(-2,f(-2))$ (derivative)

The derivative of $f(x)=x^{2}+x$ is $f^\prime(x)=2x + 1$. Evaluate at $x=-2$, $f^\prime(-2)=2\times(-2)+1=-4 + 1=-3$.

Step6: Find the equation of the tangent line

The point - slope form of a line is $y - y_1=m(x - x_1)$. We have $x_1=-2,y_1=f(-2)=2$ and $m=-3$. So $y - 2=-3(x + 2)$. Expand to get $y-2=-3x-6$, or $y=-3x - 4$.

Answer:

a) - 1
b) $h - 3$
c) - 3
d) $y=-3x - 4$