Question

find an equation of the line tangent to the graph of f(x)=x^3 at (-1,-1). the equation of the tangent line is y=□. (type an expression using x as the variable.)

Explanation:

Step1: Find derivative of $f(x)$

$f'(x)=3x^{2}$

Step2: Evaluate derivative at $x = - 1$

$f'(-1)=3(-1)^{2}=3$ (slope of tangent line)

Step3: Use point - slope form $y - y_1=m(x - x_1)$

$y-(-1)=3(x - (-1))$
$y + 1=3(x + 1)$
$y=3x+2$

Answer:

$y = 3x+2$