Equations of a tangent line

Question

Suppose that f(x)=−6x2−2x.

Find the equation of the line tangent to f(x) at X =-1

Is the answer not -12x-2 ??

Answer 1

In order to find the tangent of the the line at x=-1. We must first take the first derivative. We have that f(x) = -6x² - 2x. Then, the first derivative using the power-rule is f'(x) = -12x - 2. Then, the slope of the tangent line at x = -1 is f'(-1) = (-12)(-1) - 2 = 10. Then, we just need the original point at x=-1 by using out original equation. We have:

f(x) = (-6)(-1)² - 2(-1) = -4. Then we have our point and out slope for our tangent line: (x,y) = (-1,-4) and m=10. Then, we go back to some algebra and use y=mx+b and solve.

f(x) = mx+b

-4 = 10*(-1) + b => b = 6

Then, the equation of our tangent line is. f(x) = 10x+6

Answer 2

Garrett

12x-2 is the first part of the solution, not the final answer

1. Find the gradient of the tangent at x=-1. This is f' (-1). f'(x) = 12x-2 so f'(-1) = -14
2. When x=-1, f(x) = 8. So the tangent line has gradient -14, and passes through ( -1, 8)
3. y = mx + c. m = -14. y = -14x + c. when x=-1 y = 8. so 8 = 14 + c, so c = -6
4. So tangent line is : y = -14x - 6

Mike

1.