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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