Determine the values of a, b and c so that the graph of f(x) = ax 2 + bx + c has x-intercepts at (0,0) and (8,0) and has a tangent with a slope of 16 where x = 2.

Answer 1

Given:

f(x) = ax² + bx + c

f(0) = 0

f(8) = 0

f'(2) = 16

Find:

a = ?

b = ?

c = ?

Solution:

f(0) = 0 = a * 0 + b * 0 + c ∴ c = 0

f(8) = 0 = 64a + 8b + c ∴ 64a + 8b = 0

f'(x) = 2ax + b

f'(2) = 16 = 4a + b ∴ 4a + b = 16

So, 2 equations, 2 unknowns, 100% solvable!

64a + 8b = 0

-8 ( 4a + b ) = -8 * 16

-32a - 8b = -128

-----------------------

32a = -128

∴ a = -4

4(-4) + b = 16

-16 + b = 16

∴ b = 32

∴ f(x) = -4x² + 32x

Also: f(x) = -4 ( x² - 8x ) = -4 x ( x - 8 )

I hope this helps! Message me in the comments if you have any questions, comments, or concerns!

Answer 2

f(x) = a(x-0)(x-8) = ax(x-8) = ax^2 -8ax

f'(x) = 2ax -8a

f'(2) = 16 = 2(2)a-8a= -4a

-4a= 16

a= -4

f(x) = -4x^2 +32x

f(x) =ax^2 +bx +c, where a=-4, b=32, c=0

Answer 3

Laura,

you are given f(x) = a(x - 0)(x - 8) and you need to find the value for "a" which makes f'(2) = 16. Does this help?

Andrew F.