Question

find an equation for the line that is tangent to the curve y = x^3 - 9x at the point (-3,0)
the equation is y =
(type an expression using x as the variable.)

Explanation:

Step1: Find the derivative of the function

The derivative of $y = x^{3}-9x$ using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$ is $y'=3x^{2}-9$.

Step2: Evaluate the derivative at the given point

Substitute $x=-3$ into $y'$. So $y'(-3)=3(-3)^{2}-9=3\times9 - 9=27 - 9 = 18$. The value of the derivative at the point $(-3,0)$ is the slope $m$ of the tangent line, so $m = 18$.

Step3: Use the point - slope form of a line

The point - slope form is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(-3,0)$ and $m = 18$. Substituting these values gives $y-0 = 18(x + 3)$.

Step4: Simplify the equation

$y=18x+54$.

Answer:

$18x + 54$