SOLUTION: Find in the form y=ax^2+bx+c, the equation of the quadratic whose graph cuts the x-axis at 5, passes thrpugh (2,5) and has axis of symmetry x=1

Question 998572: Find in the form y=ax^2+bx+c, the equation of the quadratic whose graph cuts the x-axis at 5, passes thrpugh (2,5) and has axis of symmetry x=1
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Find in the form y=ax^2+bx+c, the equation of the quadratic whose graph cuts the x-axis at 5, passes through (2,5) and has axis of symmetry x=1
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x-intercept at (5,0) which is 4 to the right of the axis of symmetry.
x-intercept at (-3,0) which is 4 to the left of the axis of symmetry.
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Form:: y = a(x-5)(x+3)+c
Using (2,5), solve for "c"::
5 = a(-3)(5)+c
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Cheers,
Stan H.

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