AP AB Calculus Question

Given the function f defined by f(x) = x^3-x^2-4x+4

a. Find the zeroes of f
b. write an equation of the line tangent to the graph of f at x = 1.
c. 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.

Do not worry about answering all of them, if you answer more than another person and they are right 10 points to you. Thanks!

a) x^3 - x^2 - 4x + 4
x^2(x-1) - 4(x-1)=0
(x-1)(x^2-4)=0
(x-1)(x-2)(x+2)=0
x= 1 or x= 2 or x= -2


b) find the derivative
f'(x) = 3x^2 - 2x - 4
f'(1) = 3(1)^2 - 2(1) -4
f'(1) = -3
plug in x=1 to find y, then use point-slope form
y- 0 = -3(x-1)
y= -3x + 3

c) i just looked at the graph and could tell it is at (2,0) y= 4x - 8(equation of tangent line)

mathmatically

slope(m) = f'(x) = 3x^2 - 2x -4
y= mx+c and the point (0,-8) is on the graph so

-8= 0 + c so c=-8
the equation now is
y=mx-8
m=f'(a) =3 a^2 - 2a - 4

as (a,b) is on the graph so
b= a^3 - a^2 - 4a +4
it is also on the line so
b=(3a^2 -2a -4) a -8 = 3a^3 -2a^2 -4a -8
so
a^3 - a^2 -4a +4 = 3a^3 -2a^2 -4a -8
2a^3 -a^2 -12 =0
solve for a and 'a' is 2
so b= 0
(2,0)

f(x) = x^3-x^2-4x+4

a. Find the zeroes of f
just plug in zero for every x
f(0) = 4

b. write an equation of the line tangent to the graph of f at x = 1.
find the slope of the line, it's tangent is 1/the slope of the line
then plug it in the slope-line equation: y = mx + 1 (m=slope)

c. 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'll let you do this yourself