SOLUTION: Find a quadratic equation function f(x)=ax^2 + bx + c whose vertex is (1,2) and whose y-intercept is (0,3)

Question 1158924: Find a quadratic equation function f(x)=ax^2 + bx + c whose vertex is (1,2) and whose y-intercept is (0,3)
Found 2 solutions by Boreal, MathTherapy:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
may write as a(x-h)^2+k where( h, k) are the vertex
This is a(x-1)^2+2
or simply a(x-1)^2+2, which is x^2-2x+3, letting a=1

Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!

Find a quadratic equation function f(x)=ax^2 + bx + c whose vertex is (1,2) and whose y-intercept is (0,3)

Equation of a quadratic function:
------ Substituting (0, 3) for (x, y), and (1, 2), for (h, k), or the vertex
3 = a + 2___3 - 2 = a____1 = a
Equation:
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