Solve for a, b, and c

[fred2028]

The question in question is:

Given y ax^2 + bx + c, the points (0, 0) and (8, 0), and when x = 2 the slope of the tangent is 16, solve for a, b, and c.

So here's what I did.Knowing that (0, 0) exists, c = 0. So we're left with y = ax^2+bx. The equation can be written as:

y = ax(x-8)

as a factored form.

Now, derivative of y=ax^2 + bx is y' = 2ax + b

Substitute 2, 16

16 = 2a2 + b
16 = 4a + b

However, i do not know how to proceed from there. Help please?

 

[galactus]

You're doing OK.

You have two points (0,0) and (8,0).

\(\displaystyle a(0)^{2}+b(0)+c=0, \;\ c=0\)

\(\displaystyle a(8)^{2}+b(8)+c=0; \;\ 64a+8b=0; \;\ 8a+b=0\)

Derivative is 16 at x=2:

\(\displaystyle 2a(2)+b=16; \;\ 4a+b=16\)

There, you have two equations with two unknowns. Solve the system and you're all set.

\(\displaystyle 8a+b=0\)

\(\displaystyle 4a+b=16\)

Just use substitution or elimination to solve.

 

[fred2028]

Thanks for the quick and prompt reply! I understand it now.