# SOLUTION: I am given the zeros (ie. cross x axis at -3 and 4), and told the equation crosses the y axis at 48, how would I make a quadratic equation out of this?

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Question 157251: I am given the zeros (ie. cross x axis at -3 and 4), and told the equation crosses the y axis at 48, how would I make a quadratic equation out of this?
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I am given the zeros (ie. cross x axis at -3 and 4), and told the equation crosses the y axis at 48, how would I make a quadratic equation out of this?
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Since the problem states that the quadratic crosses the x-axis at -3 and 4 we now KNOW two factors of the quadratic. So far, then, we have:
y = (x+3)(x-4)
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Notice that to determine where a quadratic crosses the y-axis, set x=0 and solve for y:
y = (x+3)(x-4)
y = (0+3)(0-4)
y = (3)(-4)
y = -12
And, because the problem states that it "crosses the y axis at 48"...
To get it to 48, we ask ourself what do we multiply -12 with to get 48? And, the answer is -4.
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So, now we have:
y=-4(x+3)(x-4)
y=-4(x^2-x-12)
y=-4x^2+4x+48 (this is what they are looking for)
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