Question

question 3
find an equation of the tangent line to the curve y = x^3 - 4x at the point (4, 48).
o y = -44x - 4
o y = 12x + 128
o y = 44x - 128
o y = x - 48
o y = 12x - 4

Explanation:

Step1: Find the derivative of the function

The derivative of $y = x^{3}-4x$ using the power - rule $(x^n)'=nx^{n - 1}$ is $y'=3x^{2}-4$.

Step2: Evaluate the derivative at the given x - value

Substitute $x = 4$ into $y'$. So $y'(4)=3\times4^{2}-4=3\times16 - 4=48 - 4=44$. This is the slope $m$ of the tangent line.

Step3: Use the point - slope form of a line

The point - slope form is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(4,48)$ and $m = 44$.
$y-48=44(x - 4)$.

Step4: Simplify the equation

$y-48=44x-176$.
$y=44x-176 + 48$.
$y=44x-128$.

Answer:

$y = 44x-128$