How do you find the equation of the line tangent to the graph of y=x^2-2x at the point (-2,8)?

3 Answers
Aug 10, 2015

y=-6x-4

Explanation:

Take the derivative of the given function

(dy)/(dx)=2x-2

Find (dy)/(dx) for given value of x

(dy)/(dx)=2(-2)-2=-4-2=-6

Now create the point slope form of a line and the solve for y

y-8=-6(x-(-2))

y-8=-6(x+2)

y-8=-6x-12

y=-6x-4

Aug 10, 2015

y=-6x-4

Explanation:

y=x^2-2x

dy/dx=2x-2

The general equation for the tangent is given by:

y=mx+c

m=2x-2

Since x=-2 then:

m=(2xx-2)-2=-6

Since y=8:

8=(-6xx-2)+C

C=-4

So the equation of the tangent becomes:

y=-6x-4

Aug 11, 2015

You can also perform the linearization of the function.

f_T(a) = f'(a)(x-a) + f(a)
where f_T is the tangent line, f is the function at x = a, and f' is the derivative at x = a.

Thus:
f'(-2) = [2x - 2]|_(x = -2) = -6
f(-2) = [x^2 - 2x]|_(x = -2) = 8

Therefore, after taking a derivative and evaluating two functions at x = -2, without even using the value of y, you still get:

color(blue)(f_T(-2)) = -6(x+2) + 8 color(blue)(= -6x - 4)