Find an equation for the tangent line

Question

I am just having trouble on how to find an equation for the tangent line in this problem (last part):

1) Let f be defined by f(x)=3x - 3x^2 - 3, and let a be constant.

f(a) = 3a-3a^2-3 and f'(a) = 3 - 6a

An equation for the tangent line to y = f(x) at x = a is

y =

If anyone knows how to do this I would greatly appreciate it!

Answer 1

Mark is right,

but it is kind of interesting because a is an abstract value.

what my algebra computations conclude is

y=(-6a+3)x+3a²-3 is tangent line formula

let's say f(x)=-3x²+3x-3 and a=11

then y=(-63)x+360 would be tangent line to curve at (11,-333)

Answer 2

This test your knowledge of slope asnd the first derivative.

f'(a) is the slope of the tangent line at the point

(a, f(a))

Using point-slope form from Algebra 1.

(y - f(a)) = f'(a)(x - a)

Now substitute f(a) and f'(a) and convert to slope-intercept form.