Question

question
for the function $f(x)=x^{2}+11$, find the equation of the tangent line at $x = 5$.

Explanation:

Step1: Find the derivative of the function

The derivative of $f(x)=x^{2}+11$ using the power - rule $(x^n)' = nx^{n - 1}$ is $f'(x)=2x$.

Step2: Find the slope of the tangent line at $x = 5$

Substitute $x = 5$ into $f'(x)$. So $m=f'(5)=2\times5 = 10$.

Step3: Find the point on the function at $x = 5$

Substitute $x = 5$ into $f(x)$. $y=f(5)=5^{2}+11=25 + 11=36$. So the point is $(5,36)$.

Step4: Use the point - slope form of a line

The point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(5,36)$ and $m = 10$.
$y-36=10(x - 5)$
$y-36=10x-50$
$y=10x - 14$

Answer:

$y = 10x-14$