Question

find the equation of the tangent to the graph at the indicated point f(x)=X^2-6; a=-2 Y= ?

Answer 1

what is the indicated point? there isn't an "a" term shown in the function...

Answer 2

the hint says to compute the derivative algebraically

Answer 3

yes, absolutely... get the derivative... that gives you the slope of the function at ALL points... then to get the equation of a tangent line at a particular point, you can sub in the value of x at that point, solve the derivative for x to get the slope of the tangent line. Then you have a point and a slope... enough for a line equation

Answer 4

okay the derivative is 8 but how do I get Y

Answer 5

sorry sorry... derivative = 2x, not 2x -6

Answer 6

That is where I get stuck. y-f(-2)=m(x-(-2)) or y-8=?? (

Answer 7

When x = -2, the original function f(x) = x^2 - 6 -->> f(-2) = (-2)^2 - 6 = -2, so the point they are talking about is (-2, -2).

The value of the derivative at that point is f ' (x) = 2x -->> f ' (-2) = 2(-2) = -4 so the slope of the tangent line to f(x) at the point (-2, -2) is -4.

Answer 8

Okay so I still need to get the point. Do I have to graph it?

Answer 9

Do you remember how to get a line equation from a point and a slope?

Answer 10

(y- y1) = m(x-x1) is the point-slope form for a line of slope m going through point (x1,y1)

Answer 11

so it would be:

(y - (-2)) = -4 (x - (-2))