What if you had a quadratic function like

### Watch This

CK-12 Foundation: 1007S Graph Quadratic Functions in Vertex Form

### Guidance

Probably one of the best applications of the method of completing the square is using it to rewrite a quadratic function in vertex form. The vertex form of a quadratic function is

This form is very useful for graphing because it gives the vertex of the parabola explicitly. The vertex is at the point

It is also simple to find the

To find the

#### Example A

*Find the vertex, the* *intercepts and the* *intercept of the following parabolas:*

a)

b)

**Solution**

a)

Vertex: (1, 2)

To find the

The solutions are not real so there are **no** **intercepts.**

To find the

b)

To find the

To find the

To graph a parabola, we only need to know the following information:

- the vertex
- the
x− intercepts - the
y− intercept - whether the parabola turns up or down (remember that it turns up if
a>0 and down ifa<0 )

#### Example B

*Graph the parabola given by the function*

**Solution**

To find the

To find the

And since **turns up.**

Graph all the points and connect them with a smooth curve:

#### Example C

*Graph the parabola given by the function*

**Solution:**

To find the

Note: there is only one

To find the

Since **turns down.**

Graph all the points and connect them with a smooth curve:

Watch this video for help with the Examples above.

CK-12 Foundation: 1007 Graph Quadratic Functions in Vertex Form

### Vocabulary

- The vertex form of a quadratic function is

This form is very useful for graphing because it gives the vertex of the parabola explicitly. The vertex is at the point

- To find the
x− intercepts from the vertex form: just sety=0 and take the square root of both sides of the resulting equation.

- To find the
y− intercept, setx=0 and simplify.

### Guided Practice

*Graph the parabola given by the function*

**Solution:**

To find the \begin{align*}x-\end{align*}intercepts,

\begin{align*}\text{Set.} \ y = 0: & & 0 & = 4(x +2)^2-1 \\ \text{Subtract 1 from each side}: & & 1 & = 4(x +2)^2 \\ \text{Divide both sides by 4}: & & \frac{1}{4} & = (x +2)^2 \\ \text{Take the square root of both sides}: & & \frac{1}{2} & = \pm (x + 2)\\ \text{Separate}: & & & \frac{1}{2}=-(x+2) && \frac{1}{2}=x+2)\\ \text{Simplify}: & & & \underline{\underline{x = -2.5}} && \underline{\underline{x = -1.5}}\end{align*}

The \begin{align*}x-\end{align*}intercepts are \begin{align*}(-2.5, 0)\end{align*} and \begin{align*}(-1.5, 0)\end{align*}.

To find the \begin{align*}y-\end{align*}intercept,

\begin{align*}\text{Set} \ x = 0: & & y & = 4(0 +2)^2-1 \\ \text{Simplify:} & & y & = 15 \Rightarrow \underline{\underline{y = 15}} && y- \text{intercept:}(0, 15)\end{align*}

Since \begin{align*}a < 0\end{align*}, the parabola **turns up.**

Graph all the points and connect them with a smooth curve:

### Practice

Rewrite each quadratic function in vertex form.

- \begin{align*} y= x^2 - 6x\end{align*}
- \begin{align*}y + 1 = -2x^2 -x\end{align*}
- \begin{align*}y = 9x^2 + 3x - 10\end{align*}
- \begin{align*}y = -32x^2 + 60x + 10\end{align*}

For each parabola, find the vertex; the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}intercepts; and if it turns up or down. Then graph the parabola.

- \begin{align*}y - 4 = x^2 + 8x\end{align*}
- \begin{align*}y = -4x^2 + 20x - 24\end{align*}
- \begin{align*}y = 3x^2 + 15x\end{align*}
- \begin{align*}y + 6 = -x^2 + x\end{align*}
- \begin{align*}x^2-10x+25=9\end{align*}
- \begin{align*}x^2+18x+81=1\end{align*}
- \begin{align*}4x^2-12x+9=16\end{align*}
- \begin{align*}x^2+14x+49=3\end{align*}
- \begin{align*}4x^2-20x+25=9\end{align*}
- \begin{align*}x^2+8x+16=25\end{align*}