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:
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(x-4)(x-2) = 0
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Foil tells us the equation is:
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x^2 -2x -4x +8 = 0
x^2 -6x + 8 = 0
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So, when x=0, y = 8.
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In contrast, a y-intercept at -16 tells us the point is (0,-16).
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Let's look at the graph of x^2-6*x+8.
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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?
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-2(x-4)(x-2) = 0
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That seems to do it.
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Done.
Answer by josmiceli(19441)  (Show Source):
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