A quadratic function has its vertex at the point 11 The func

A quadratic function has its vertex at the point (1,1). The function passes through the point (2,4). Find the quadratic and linear coefficients and the constant term of the function\'

Solution

Use the vertex formula:
y = a(x - p)^2 + q.

y = a(x - 1)^2 + 1.

Now all you need is a. Plug in the point (-2,-4) to get:
- 4 = a(-2 - 1)^2 + 1
- 5 = a(-3)^2

Then, a = -5/9

Final equation is y = - (5/9)(x - 1)^2 + 1.
I\'ll let you square it and put it into standard form for the co-efficients.

y = - 5/9*x^2 + 10/9*x - 5/9 +1

y = - 5x^2/9 + 10x/9 - 4/9