Question

Let f(x) = 3 x^2 - 8.
Find an equation of the line tangent to the graph of f at the point where x = -3.

y =

Answer 1

y= -18?

Answer 2

y'=6x
6(-3)=y=-18

Answer 3

that's the slope, can you figure out the rest?

Answer 4

plug -3 into the original equation, to find the corresponding y value, then use the (x,y) point you get with the slope m=-18 with the point slope formula y-y1=m(x-x1) to solve for the line

Answer 5

You have to find the point that corresponds to x = -3 on your graph f(x)
We get: f(-3) = 3(-3)^2 -8 = 27 - 8 =19
Then you have to find the derivative of the function, to find the slope of the point at x = -3
Taking the derivative, f'(x) = 3*2x - 0
Then, f'(x) = 6x
To find the slope at point x= -3, substitute x = -3 in.
We get f'(-3) = -18
Now we now the point on f(x), which is (-3, 19), and the slope at x= -3, which is -18
Now we use the equation of the line formula which is
y-y0 = m(x-x0),
Now we just substitute our values in and get:
y-19 = -18(x--3)
y-19 = -18(x+3)