Questions: Find an equation for the tangent line to the graph of y=(x^3-4x)^10 at the point (-2,0). The equation of the tangent line is y=

Transcript text: Find an equation for the tangent line to the graph of $y=\left(x^{3}-4 x\right)^{10}$ at the point $(-2,0)$. The equation of the tangent line is $y=$ $\square$ (Simplify your answer.)

Solution

Solution Steps

Step 1: Define the Function and Compute its Derivative

We are given the function $ y = (x^3 - 4x)^{10} $. To find the equation of the tangent line at the point $(-2, 0)$, we first need to compute the derivative of the function, $ y' $, which represents the slope of the tangent line at any point $ x $.

The derivative of the function is: \[ y' = \frac{d}{dx} \left( (x^3 - 4x)^{10} \right) = 10(x^3 - 4x)^9 \cdot (3x^2 - 4) \]

Step 2: Evaluate the Derivative at the Given Point

Next, we evaluate the derivative at $ x = -2 $ to find the slope of the tangent line at the point $(-2, 0)$.

Substituting $ x = -2 $ into the derivative: \[ y'(-2) = 10((-2)^3 - 4(-2))^9 \cdot (3(-2)^2 - 4) = 10(0)^9 \cdot 8 = 0 \]

The slope of the tangent line at $(-2, 0)$ is $ 0 $.

Step 3: Write the Equation of the Tangent Line

The equation of a tangent line at a point $(x_1, y_1)$ with slope $ m $ is given by: \[ y - y_1 = m(x - x_1) \]

Substituting $ m = 0 $, $ x_1 = -2 $, and $ y_1 = 0 $: \[ y - 0 = 0(x + 2) \]

Simplifying, we find: \[ y = 0 \]