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

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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) About Me  (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
graph%28300%2C300%2C-10%2C10%2C-10%2C10%2Cx%5E2-2x%2B3%29

Answer by MathTherapy(10552) About Me  (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: matrix%281%2C3%2C+%22f%28x%29%2Fy%22%2C+%22=%22%2C+a%28x+-+h%29%5E2+%2B+k%29

matrix%281%2C3%2C+3%2C+%22=%22%2C+a%280+-+1%29%5E2+%2B+2%29 ------ Substituting (0, 3) for (x, y), and (1, 2), for (h, k), or the vertex

3 = a + 2___3 - 2 = a____1 = a

Equation: