Question

Please..I'm desperate. I have a D in this class..

Answer 1

Step-by-step explanation:

Recall that the first derivative of an equation gives us the SLOPE for the tangent line.

So first, we should find the first derivative.

$\frac{dy}{dx} = 6 - 3x^2$

With the first derivative, we can solve for the SLOPE of the tangent line at (1,5).

Using x = 1.

$slope = 6 - 3(1)^2$

A tangent line is a line. And has the form $y = mx +b$

Where m is the SLOPE of the line. and b is the y-intercept.

To get to this equation, let's use the point-slope technique.

$y-y_1 = m(x-x_1)$

$y - 5 = m(x-1)$

Solve for slope, substitute it into this equation for m and solve. That is the equation of the tangent line at the point (1,5).