The equation of the tangent to the curve y= x^3 + ax^2+ bx + 4 at the point where x = 2 is y = x - 4.
Find the values of "a" and "b".
So what I did was differentiated
y' = 3x^2 + 2ax + b
when x = 2,
y = 2^3 + 2^2a + 2b + 4
= 12 + 4a + 2b
and
y' = 3*2^2 + 2*2a + b
= 12 + 4a + b
What you need to to do know is
y - y1 = m(x - x1)
And what I know is that when you plug the "x" value into the first equation is that it will give you the y-value, and if you put the x-value into the differentiated equation, it will give you the gradient.
So this is what I did:
y = y1 and y' = m, so
y - (12 + 4a +2b) = (12 + 4a + b)*(x-2)
and that got me,
y - 12 - 4a - 2b = 4ax - 8a + bx - 2b + 12x - 24
simplify that,
y + 12 + 4a = 4ax + bx + 12x
And now I'm stuck.