find the general form of the equation of the quadratic function

Question

Use the vertex (h, k)

and a point on the graph (x, y)

to find the general form of the equation of the quadratic function.

(h, k) = (−4, −1),    (x, y) = (−6, 3)

Answer 1

Start with the vertex form because you are given the parabola's vertex, (h.k) = (-4,-1):

y = a·(x-h)² + k

y = a·(x-(-4))² + (-1)

y = a·(x+4)² - 1

To find the value of the constant a, plug in the given point (-6,3):

y = a·(x+4)² - 1

3 = a·(-6+4)² - 1

3 = a·(-2)² - 1

3 = 4a - 1

4 = 4a

1 = a

So the final equation in vertex form is:

y = (x+4)² - 1

Multiply the right hand side out to put it into standard form, y = ax² + bx + c.

Answer 2

The standard vertex form of a quadratic function is:

y = A(x-h)² + k where (h, k) is the vertex and A is the stretch factor.

First plug in the vertex:

y = A(x+4)² -1

Then plug in the point to solve for A

3 = A(-6+4)² - 1

4 = A(4)

A = 1

The equation is y = (x+4)² -1 = x² + 8x + 15