Find the equation of the tangent to the function g(x)=(x-4)^2 at its y-int.
Thanks!
the y-intercept is at (0,16)
g'(x) = 2(x-4) so g'(0) = -8
Now we have a point and a slope, so the equation of the line is
y-16 = -8(x-0)
or
y = -8x+16
Thanks!
g'(x) = 2(x-4) so g'(0) = -8
Now we have a point and a slope, so the equation of the line is
y-16 = -8(x-0)
or
y = -8x+16