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

Answer:

$y = \frac{1}{2}x^{2} - 3x + 11.5$

Step-by-step 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;

$9 = a( (-1 ) + 3)^2 + 7\\9 = a (2)^{2} + 7\\9 = 4a + 7\\-7 -7\\2 = 4a\\\frac{1}{2} = a$

Hence the equation in vertex form is;

$y = \frac{1}{2}(x - 3)^{2} + 7$

In standard form it is;

$y = \frac{1}{2}x^{2} - 3x + 11.5$