Question

for the function $f(x)=x^{2}-7$, find the equation of the tangent line at $x = - 10$.

Explanation:

Step1: Find the derivative of the function

The derivative of $f(x)=x^{2}-7$ using the power - rule $(x^n)^\prime=nx^{n - 1}$ is $f^\prime(x)=2x$.

Step2: Find the slope of the tangent line at $x = - 10$

Substitute $x=-10$ into $f^\prime(x)$. So, $m = f^\prime(-10)=2\times(-10)=-20$.

Step3: Find the point on the function at $x = - 10$

Substitute $x=-10$ into $f(x)$. $y=f(-10)=(-10)^{2}-7=100 - 7 = 93$. The point is $(-10,93)$.

Step4: Use the point - slope form of a line

The point - slope form is $y - y_1=m(x - x_1)$. Here, $x_1=-10$, $y_1 = 93$ and $m=-20$. So, $y - 93=-20(x+10)$.

Step5: Simplify the equation

$y-93=-20x - 200$. Then $y=-20x-200 + 93=-20x-107$.

Answer:

$y=-20x - 107$