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Write the equation of the quadratic function with roots 0 and 2 and a vertex at (1, 5).

User Dejal
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1 Answer

1 vote

The "standard" parabola with roots 0 and 2 is


y = (x-0)(x-2) = x^2-2x

All multiples of this parabola, i.e.


y=a(x^2-2x)

have the same roots. We can choose the factor such that the parabola passes through the desided point: if we plug 1, 5 for x, y we have


5 = a(1-2) \iff -a=5 \iff a=-5

So, our claim is that the parabola


y=-5(x^2-2x) = -5x^2+10x

has roots 0 and 2 and vertex at (1, 5).

You can easily verify this: the roots are guaranteed by the fact that we can write the equation as


y = -5x(x-2)

The vertex must be at x=1, because it's the midpoint of the roots. Moreover, if we evaluate the function at x=1 we have


y(1) = -5\cdot 1 \cdot (1-2) = -5 \cdot 1 \cdot (-1) = 5

as required.