The equation of the tangent line is
y = mx -8,
where m is the slope at the tangent point (a,b).
At (a,b), the value of y is
a^3 - a^2 + 4a + 4 = b
and the slope is
dy/dx @ x=a =
3x^2 -2x -4 = 3a^2 -2a -4 = m
From the tangent line equation, we also know that, at (a,b)
b = m a -8
So there are 3 equations in the three unknowns, m, a and b.
It is a bit messy, but that is as far as I can go.
The function f is defined by
f(x) = x^3 - x^2 - 4x + 4
The point (a,b) is on the graph of f and the line tangent to the graph at (a,b) passes through the point (0, -8) which is not on the graph of f. Find the values of a and b.
I have no clue how to solve this
I know the derivative of f prime (x) is
3x^2 - 2x - 4 and so (a,b) would be on the derivative line and so would (0, -8) but how do i find (a,b)???
1 answer
1 answer