Question

The quadratic function which takes the value 41 at x= -2 and the value 20 at x=5 and minimized at x=2 is y=Ax^2 - Bx + C. The minimum value of this function is D . What are the values of A,B,C,D?

Answer 1

This is a parabola, opening upward (or it wouldn't have a minimum value). Written in the form \(y=ax^2+bx+c\) we can find the vertex x coordinate at x=-b/2a. We also know that that x coordinate is x=2. Combined with the value of the function being 20 at x=5 and 41 at x=-2 we can solve to find A B and C. Once you're that, evaluating the function at x=2 gives you the value of D

Answer 2

well i tryed to solve fot 3 times i couldnt let one more go again

Answer 3

@bhusal did it work any better?

Answer 4

Let's see what we know:
at \(x = -2,\)
\[f(x) = 41 = Ax^2-Bx + C = A(-2)^2-B(-2) + C = 4A + 2B + C\]
at \(x = 5,\)\[f(x) = 20 = Ax^2-Bx + C = A(5)^2 -B(5) + C = 25A-5B+C\]giving us two equations in three unknowns:
\[4A+2B+C=41\]\[25A-5B+C=20\]If we combine them by subtraction,
\[ ~~~~ 4A+2B+C=41\]\[~~25A-5B+C=20\]---------------------\[-21A+7B=21\]

Answer 5

Now we also know that this is a parabola, and written in the form \(y = ax^2+bx+c,\) with \(a = A, b=-B, c=C\) so the vertex (which will be the minimum point on the parabola) can be found at \(x = -b/2a\). We also know that the minimum point is at \(x = 2\) giving us a relationship between \(A\) and \(B\):
\[x = 2 = -\frac{b}{2a} = -\frac{(-B)}{2A}\]or\[2 = \frac{B}{2A}\rightarrow B=4A\]Now we substitute that in our equation \(-21A+7B=21\)
\[-21A + 7(4A) = 21\]and solve for \(A\).
After doing so, use \(B=4A\) to find the value of \(B\).
Then use the values of \(A\) and \(B\) in \(4A+2B+C=41\) or \(25A-5B+C=20\) to find the value of\(C\).
FInally, evaluate the polynomial at \(x = 2\) to find the value of \(D\).