SOLUTION: A quadratic function has zeros at x=2 and x=4 and a y-intercept at -16. What is the quadratic function in standard form? I know I have to unfoil it, but I don't see how it is po

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Question 513375: A quadratic function has zeros at x=2 and x=4 and a y-intercept at -16. What is the quadratic function in standard form?
I know I have to unfoil it, but I don't see how it is possible. If y=-16, wouldn't one of zeros have to be negative?

Found 2 solutions by oberobic, josmiceli:
Answer by oberobic(2304)   (Show Source): You can put this solution on YOUR website!
Unfoil tells us the equation is:
.
(x-4)(x-2) = 0
.
Foil tells us the equation is:
.
x^2 -2x -4x +8 = 0
x^2 -6x + 8 = 0
.
So, when x=0, y = 8.
.
In contrast, a y-intercept at -16 tells us the point is (0,-16).
.
Let's look at the graph of x^2-6*x+8.
.

.
Hmmm...
So, perhaps the graph is going the other way?
And to have a y-intercept of (0,-16), we perhaps need to multiply by the constant -2?
.
-2(x-4)(x-2) = 0

.
That seems to do it.
.
Done.

Answer by josmiceli(19441)   (Show Source): You can put this solution on YOUR website!
The general form is

The zeros:


and


-------------------
When ,


-----------
Now I have


(1)
and


(2)
Subtract (1) from (2)
(2)
(1)


and
(2)
(2)
(2)
The equation is:

check:
(2,0)



(4,0)



OK


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