a. use the definition (m_{tan}=lim_{x\rightarrow a}\frac{f(x)-f(a)}{x - a}) to find the slope of the line…

a. use the definition (m_{tan}=lim_{x\rightarrow a}\frac{f(x)-f(a)}{x - a}) to find the slope of the line tangent to the graph of (f) at (p). when using the definition (lim_{x\rightarrow a}\frac{f(x)-f(a)}{x - a}), which is an expression for the limit to use to find the slope of the tangent line at (p)? a. (lim_{x\rightarrow - 1}\frac{\frac{-16}{x^{2}+8}}{x - 1}) b. (lim_{x\rightarrow - 1}\frac{\frac{-8}{x^{2}+8}}{x - 1}) c. (lim_{x\rightarrow - 1}\frac{\frac{-8}{x^{2}+8}}{x + 1}) d. (lim_{x\rightarrow - 1}\frac{\frac{-8}{x^{2}-8}}{x + 1}) b. determine an equation of the tangent line at (p). (f(x)=\frac{-8}{x^{2}},p(-1,-8)) c. plot the graph of (f) and the tangent line at (p). (m_{tan}=square) (simplify your answer.)

a. use the definition (m_{tan}=lim_{x\rightarrow a}\frac{f(x)-f(a)}{x - a}) to find the slope of the line tangent to the graph of (f) at (p). when using the definition (lim_{x\rightarrow a}\frac{f(x)-f(a)}{x - a}), which is an expression for the limit to use to find the slope of the tangent line at (p)? a. (lim_{x\rightarrow - 1}\frac{\frac{-16}{x^{2}+8}}{x - 1}) b. (lim_{x\rightarrow - 1}\frac{\frac{-8}{x^{2}+8}}{x - 1}) c. (lim_{x\rightarrow - 1}\frac{\frac{-8}{x^{2}+8}}{x + 1}) d. (lim_{x\rightarrow - 1}\frac{\frac{-8}{x^{2}-8}}{x + 1}) b. determine an equation of the tangent line at (p). (f(x)=\frac{-8}{x^{2}},p(-1,-8)) c. plot the graph of (f) and the tangent line at (p). (m_{tan}=square) (simplify your answer.)

Answer

Explanation:

Step1: Recall the slope - of - tangent formula

The slope of the tangent line to the graph of $y = f(x)$ at the point $x=a$ is given by $m_{tan}=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x - a}$. Here, $f(x)=-\frac{8}{x^{2}}$, $a=-1$, and $f(-1)=-\frac{8}{(-1)^{2}}=-8$.

Step2: Substitute into the formula

We have $\lim_{x\rightarrow - 1}\frac{-\frac{8}{x^{2}}-(-8)}{x - (-1)}=\lim_{x\rightarrow - 1}\frac{-\frac{8}{x^{2}} + 8}{x + 1}=\lim_{x\rightarrow - 1}\frac{\frac{-8 + 8x^{2}}{x^{2}}}{x + 1}=\lim_{x\rightarrow - 1}\frac{8(x^{2}-1)}{x^{2}(x + 1)}$.

Step3: Factor the numerator

Since $x^{2}-1=(x + 1)(x - 1)$, the limit becomes $\lim_{x\rightarrow - 1}\frac{8(x + 1)(x - 1)}{x^{2}(x + 1)}$.

Step4: Cancel out the common factor

Cancel out the common factor $(x + 1)$ (for $x\neq - 1$), we get $\lim_{x\rightarrow - 1}\frac{8(x - 1)}{x^{2}}$.

Step5: Evaluate the limit

Substitute $x=-1$ into $\frac{8(x - 1)}{x^{2}}$, we have $\frac{8(-1 - 1)}{(-1)^{2}}=\frac{8\times(-2)}{1}=-16$.

Answer:

The slope of the tangent line $m_{tan}=-16$