SOLUTION: How do i prove values of a=1/2 and b=2, if given f(x)=-ax^2+bx+c and the tangent to the graph of f at the point (-1;7/2) is 3

Algebra ->  Square-cubic-other-roots -> SOLUTION: How do i prove values of a=1/2 and b=2, if given f(x)=-ax^2+bx+c and the tangent to the graph of f at the point (-1;7/2) is 3      Log On


   

Question 1111541: How do i prove values of a=1/2 and b=2, if given f(x)=-ax^2+bx+c and the tangent to the graph of f at the point (-1;7/2) is 3
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Something is missing. Is there a typo?
What was meant by "the tangent to the graph of f at the point (-1;7/2) is 3" is not obvious to me.
Is it that 3 is the slope of the tangent?
It could not be that y=3 is the equation
of the straight line tangent to the graph at the point (-1,7/2),
because that line must contain the point (-1,7/2).

We are told f%28x%29=-ax%5E2%2Bbx%2Bc , and f%28-1%29=-a%28-1%29%5E2%2Bb%28-1%29%2Bc=-a-b%2Bc=7%2F2 .
The slope of the tangent to f%28x%29=-ax%5E2%2Bbx%2Bc at a point with any x is
df%2Fdx=-2ax%2Bb , the derivative of f%28x%29 .
When x=-1 , the value of the derivative is
-2a%28-1%29%2Bb=2a%2Bb .

If that slope is 3 ,
we have system%28-a-b%2Bc=7%2F2%2C2a%2Bb=3%29 .
That system of equations has infinite solutions,
so there is no way to prove that a=1/2 and b=2 with
Knowing just that the tangent at (-1,7/2) has a slope of 3.
red%28system%28a=1%2F2%2Cb=2%2Cc=6%29%29 is one of them, but green%28system%28a=1%2Cb=1%2Cc=11%2F2%29%29 and blue%28system%28a=3%2F2%2Cb=0%2Cc=5%29%29
are also among the infinite number of solutions:


Another piece of information would give us another equation,
which could complete a system of equation with a unique solution.