College of St Joseph The Vertex Form of a Parabola Questions

Choose two distinct solutions to a quadratic equation and workbackwards to write the original equation for the parabola in vertexform.

Vertex form:

$y=a{(x?h)}^2+k$

There are a few different ways you can do this! I’m going to walkyou through the method that students often find more difficult.

EXAMPLE:

My two chosen solutions are x = 4 and x = -3
So in working backwards, I have

$y=(x?4)(x+3)$

$y=x^2?x?12$

$y=x^2?x+\frac{1}{4}?12?\frac{1}{4}$

$y=(x?\frac{1}{2})(x?\frac{1}{2})?\frac{48}{4}?\frac{1}{4}$

$y={(x?\frac{1}{2})}^2?\frac{49}{4}$

The other method is to multiply using FOIL until you get your parabola in standard form:

$y=a{x}^2+bx+c$

Then find the vertex (h,k):

The h value of the vertex is $\frac{?b}{2a}$

The k value of the vertex is found by plugging in $\frac{?b}{2a}$ for x and simplifying to find y

Once you find the vertex of (h,k), use it to fill in the following:

$y={(x?h)}^2+k$

EXAMPLE:

My two chosen solutions are x = 4 and x = -3

$y=(x?4)(x+3)$

$y=x^2?x?12$

$\frac{?b}{2a}=\frac{1}{2\left(1\right)}=\frac{1}{2}$

$y={\left(\frac{1}{2}\right)}^2?\frac{1}{2}?12$

$y=\frac{1}{4}?\frac{1}{2}?12$

$y=\frac{1}{4}?\frac{2}{4}?\frac{48}{4}$

$y=?\frac{49}{4}$

So

$y={(x?\frac{1}{2})}^2?\frac{49}{4}$