Question

last question for tonight

Let f(x) = 3 x^2 - 11.

Find an equation of the line tangent to the graph of f at the point where x =-2

y=

Answer 1

i know i find the slope

Answer 2

dy/dx = 6x
x=-2,
dy/dx= -12

Answer 3

y-1=-12(x-(-2))
y=-12x-23

Answer 4

Anything not sure just ask.

Answer 5

ok then thanks im studying problems in my online hw for a test friday

Answer 6

f(x) = 3 x^2 - 11.

f( -2) = 3 * 4 - 11 = 1
f' (x) = 6x
slope m = f'( -2) = 6(-2) = -12
=> Tangent line y -1 = -12 ( x +2)
y = -12x -23