Question

Write the standard form of the quadratic function whose graph is a parabola with vertex (-3,5) passes through point (0,14)

Answer 1

The vertex form of a parabola equation is the following:

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

Where (h, k) is the vertex of the parabola. The vertex is (h, k) = (-3, 5). Replacing in the formula:

`y=a(x+3)^2+5`

To determine the value of "a" we replace the point (x, y) = (0, 14). Replacing we get:

`14=a(0+3)^2+5`

We solve for "a" first by subtracting 5 to both sides:

`14-5=a(3)^2`

Solving the operations:

`9=9a`

Dividing both sides by 9 we get:

`\begin{gathered} (9)/(9)=a \\ 1=a \end{gathered}`

Replacing the value of "a":

`y=(x+3)^2+5`

The standard form of the parabola equation is:

`y=ax^2+bx+c`

we can take the given equation by solving the square:

`y=x^2+6x+9+5`

Solving the operations:

`y=x^2+6x+14`