Question

HELP PLEASE! Write a quadratic function that passes through the point (-1,9), has an axis of symmetry of x=-3 and a minimum value of 7.

Answer 1

`y = (1)/(2)x^(2) - 3x + 11.5`

Explanation:

Vertex form of a quadratic equation;

`y = a( x - h )^(2) + k`

Vertex of the parabolas (h, k)

The vertex of the parabola is either the minimum or maximum of the parabola. The axis of symmetry goes through the x-coordinate of the vertex, hence h = -3. The minimum of the parabola is the y-coordinate of the vertex, so k= 7. Now substitute it into the formula;

`y = a ( x + 3 ) ^(2) + 7`

Now substitute in the given point; ( -1, 9) and solve for a;

`9 = a( (-1 ) + 3)^2 + 7\\9 = a (2)^(2) + 7\\9 = 4a + 7\\-7 -7\\2 = 4a\\(1)/(2) = a\\`

Hence the equation in vertex form is;

`y = (1)/(2)(x - 3)^(2) + 7`

In standard form it is;

`y = (1)/(2)x^(2) - 3x + 11.5`