PLEASE HELP! Pre-Calc question

[Samanthalovesmath]

How would I go about solving

 

[HallsofIvy]

I thought I had answered this before. Perhaps it was on a different board. If y is a quadratic function of x then it can be written $\displaystyle y= a(x- b)^2+ c$. When x= b, y= a(b- b)^2+ c= c so (b, c) is a point on the graph. Since a square is never negative, y is always c plus something so higher than c (if a is positive) or alway c minus something so lower than c (if a is negative). Here the highest point (the "vertex" of the parabola) is (-4, 4) so b= -4 and c= 4. We can write the equation as $\displaystyle y= a(x+ 4)^2+ 4$. We also told that (0, -12) is on the parabola so $\displaystyle -12= a(0+ 4)^2+ 4$. -12= 16a+ 4. Solve that for a.

 

[JeffM]

I do not like this problem at all.

In general, it takes three points to determine a quadratic. Here you are given just two points. Consequently we must guess that the point (-4, 4) is supposed to be a maximum ALTHOUGH THE IDIOT WHO WROTE THE PROBLEM DOES NOT DISCLOSE THAT.

The general form of a quadratic is [MATH]ax^2 + bx + c.[/MATH]
We are given the point (0, - 12).

[MATH]\therefore a * 0^2 + b * 0 + c = - 12 \implies c = - 12.[/MATH]
The graph shows that the parabola has a maximum so a is negative. Furthermore, the extremum of a parabola occurs when x = -(b/2a).

Assuming that the maximum is when x = - 4, then

[MATH]-4 = - \dfrac{b}{2a} \implies b = 8a.[/MATH]
[MATH]a(-4)^2 + 8a(- 4) - 12 = 4 \implies 16a - 32a = 16 \implies a = -1 \implies b = - 8.[/MATH]
Let's check.

[MATH]f(x) =-x^2 - 8x - 12 \implies \\ f(0) = - 12 \ \checkmark\\ - \dfrac{- 8 }{2(-1)} = - \dfrac{8}{2} = - 4 \ \checkmark \\ f(-4) = -1(-4)^2 - 8(-4) - 12 = - 16 + 32 - 12 = 32 - 28 = 4 \ \checkmark[/MATH]