Question

ind the equation of the tangent line to the graph of the given function at the point with the indicated x-coordinate. f(x) = x3; x = −2

Answer 1

To find the equation of the tangent line at x = -2 for the function
`f(x) = x^3`, calculate the derivative, evaluate it at x = -2 to get the slope, and use the point-slope form with the point (-2, -8) to obtain the equation y = 12x + 16.

Step-by-step explanation:

To find the equation of the tangent line to the graph of the function
`f(x) = x^3` at the point with the x-coordinate x = -2, you need to follow these steps:

1. Calculate the derivative of f(x) to find the slope of the tangent line at any point.
2. Find the slope at x = -2 by plugging it into the derivative.
3. Use the point-slope form of the equation for a straight line to write the equation of the tangent line.

Let's go through the steps:

1. The derivative of
   `f(x) = x^3` is
   `f'(x) = 3x^2`.
2. Substituting x = -2 into f'(x) gives the slope:
   `f'(-2) = 3(-2)^2 = 12`.
3. The point on the curve is
   `(-2, (-2)^3)` or (-2, -8). Now we use the point-slope form: y - y1 = m(x - x1), which gives us y + 8 = 12(x + 2).

Therefore, the equation of the tangent line is y = 12x + 16.