Question

question given $f(x)=x^{2}-4$, find the equation of the tangent line of $f$ at the point where $x = - 4$.

Explanation:

Step1: Find the derivative of the function

The derivative of $f(x)=x^{2}-4$ using the power - rule $(x^n)'=nx^{n - 1}$ is $f'(x)=2x$.

Step2: Find the slope of the tangent line at $x = - 4$

Substitute $x=-4$ into $f'(x)$. So $m=f'(-4)=2\times(-4)=-8$.

Step3: Find the y - coordinate of the point on the function at $x=-4$

Substitute $x = - 4$ into $f(x)$. $y=f(-4)=(-4)^{2}-4=16 - 4=12$.

Step4: Use the point - slope form of a line

The point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(-4,12)$ and $m=-8$.
$y - 12=-8(x + 4)$.
Expand to get $y-12=-8x-32$.
Then $y=-8x - 20$.

Answer:

$y=-8x - 20$