Question

find an equation for the tangent line to the graph of the given function at (-5,23)
f(x)=x^{2}-2
find an equation for the tangent line to the graph of f(x)=x^{2}-2 at (-5,23)
y = (type an expression using x as the variable )

Explanation:

Step1: Find the derivative of the function

The derivative of $f(x)=x^{2}-2$ 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 $m=f'(-5)=2\times(-5)=-10$.

Step3: Use the point - slope form of a line

The point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(-5,23)$ and $m=-10$.
$y - 23=-10(x + 5)$.

Step4: Simplify the equation

$y-23=-10x-50$.
$y=-10x - 50 + 23$.
$y=-10x-27$.

Answer:

$y=-10x - 27$