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