Problema Solution

Find the formula for a quadratic function that passes through the points (2,5) and (4,33) and has its axis of symmetry x=2

Answer provided by our tutors

For a quadratic function in standard form, y=ax^2+bx+c, the axis of symmetry is a vertical line x=−b/2a.

So we know that -b/2a = 2 or if we multiply both sides by (-2a) we get b = -4a.

y=ax^2+bx+c goes through (2,5) and (4,33) means:

a*2^2 + 2b + c = 5

a*4^2 + 4b + c = 33

If we plug b = -4a into the above equations we get:

a*2^2 + 2*(-4a) + c = 5

a*4^2 + 4*(-4a) + c = 33

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click here to see the above system of equations solved for a and c

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a = 7

c = 33

b = (-4)*7

b = -28

The quadratic equation is:

y = 7x^2 - 28x + 33