Given the function f(x)=ax^2+bx+c, we are given that it has x-intercepts at (0,0) and (8,0) and a tangent with slope=16 at the point x=2. Find the value of a,b, and c.

Firstly, to begin this problem we must think of the best way to substitute the given values into the formula. For a start, we are given the tangent at x=2. In other words, this is equal to the derivative at x=2. Therefore, if we calculate f’(x), we find that f’(x)=2ax+b. Substituting x=2 and f’(2)=16, we have that 16=2(2)a+b=> b=16-4aWe are also given that there is an x intercept at (0,0), and thus:f(0)=a(0)²+b(0)+c=> 0=cMoreover, we know the other x-intercept is (8,0):0=a(8)²+b(8)+0=> 64a+8b=0.
We now have a system of equations with two variables a and b.
64a+8b=04a+b=16.
Solving for b in the second equation and substituting into the first, b=16-4aand 64a+8(16-4a)=0=> 64a-32a+128=0=>32a=-128=>a=-4
Lastly. To find b, we substitute the value of a into one of the equations,b=16-4(-4)=> b= 32
Finally, now that we have found a=-4,b=32, and c=0,f(x)=-4x²+ 32x