Find an equation of the tangent line to the curve at a give point: -2x^3 - 5x (-1,3)

Question

I know you can use (y-y1) = m(x-x1) to solve but I’m not sure how to apply it exactly.

Answer 1

f(x) = -2x³ - 5x

Find first derivative of f(x):

f'(x) = -6x² - 5

Plug x value into f'(x) to find the slope at x:

m = f'(-1) = -6(-1)² - 5 = -11

Use the point-slope formula to solve for the y-intercept (b):

The only thing that doesn't make sense is the original equation doesn't go through the point (-1,3). It would go through the point (-1,7), so I'll use this as the point.

y = mx + b

7 = -11(-1) + b

b = -4

Equation of the tangent line: y = -11x - 4

Answer 2

-2x^3-5x

take the derivative

the derivative is the slope of the tangent line

--6x^2 -5

x=-1 so

-6(-1)^2-5=`11

slope of tangent line is -11

y-3 = -11(x+1)

y=-11x-11+3

11x+y=-8

(-1,3) is on the cubic equation -2x^3-5x

-2(-1)^3 -5 (-1) = -2+5=3