138. find the equation of the tangent line to the graph of $f(x)=x^{2}+\frac{4}{x}-10$ at $x = 8$.

Question

Accepted Answer

# Explanation:
## Step1: Find the value of $ f(8) $
Substitute $ x = 8 $ into $ f(x)=x^{2}+\frac{4}{x}-10 $.
$$
\begin{align*}
f(8)&=8^{2}+\frac{4}{8}-10\
&=64 + 0.5-10\
&=54.5=\frac{109}{2}
\end{align*}
$$
## Step2: Find the derivative $ f^{\prime}(x) $
Using the power rule $ (x^{n})^\prime=nx^{n - 1} $ and $ (\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}} $ (where $ u = 4$, $ v=x $), $ f^{\prime}(x)=2x-\frac{4}{x^{2}} $.
## Step3: Find the value of $ f^{\prime}(8) $
Substitute $ x = 8 $ into $ f^{\prime}(x)=2x-\frac{4}{x^{2}} $.
$$
\begin{align*}
f^{\prime}(8)&=2\times8-\frac{4}{8^{2}}\
&=16-\frac{4}{64}\
&=16-\frac{1}{16}\
&=\frac{256 - 1}{16}=\frac{255}{16}
\end{align*}
$$
## Step4: Use the point - slope form $ y - y_{1}=m(x - x_{1}) $
Here $ x_{1}=8 $, $ y_{1}=f(8)=\frac{109}{2} $, $ m = f^{\prime}(8)=\frac{255}{16} $.
$$
\begin{align*}
y-\frac{109}{2}&=\frac{255}{16}(x - 8)\
y-\frac{109}{2}&=\frac{255}{16}x-\frac{255\times8}{16}\
y-\frac{109}{2}&=\frac{255}{16}x-\frac{255}{2}\
y&=\frac{255}{16}x-\frac{255}{2}+\frac{109}{2}\
y&=\frac{255}{16}x-\frac{255 - 109}{2}\
y&=\frac{255}{16}x-\frac{146}{2}\
y&=\frac{255}{16}x - 73
\end{align*}
$$

# Answer:
$ y=\frac{255}{16}x - 73 $

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QUESTION IMAGE

Question

Explanation:

Step1: Find the value of \( f(8) \)

Step2: Find the derivative \( f^{\prime}(x) \)

Step3: Find the value of \( f^{\prime}(8) \)

Step4: Use the point - slope form \( y - y_{1}=m(x - x_{1}) \)

Answer: