Suppose \begin{align*}p(m)=\frac {1}{12}m^2 - m + 5\end{align*}

### Graphs of Quadratic Functions

Previous Concepts introduced the concept of factoring quadratic trinomials of the form \begin{align*}0=ax^2+bx+c\end{align*}**standard form for a quadratic equation.** The most basic quadratic equation is \begin{align*}y=x^2\end{align*}*quadrare,* meaning “to square.” By creating a table of values and graphing the ordered pairs, you find that a quadratic equation makes a \begin{align*}U\end{align*}**parabola.**

#### Example A

Graph the most basic quadratic equation, \begin{align*}y=x^2\end{align*}

\begin{align*}x\end{align*} |
\begin{align*}y\end{align*} |
---|---|

–2 | 4 |

–1 | 1 |

0 | 0 |

1 | 1 |

2 | 4 |

**Solution:**

By graphing the points in the table, you can see that the shape is approximately like the graph below. This shape is called a **parabola.**

**The Anatomy of a Parabola**

A parabola can be divided in half by a vertical line. Because of this, parabolas have **symmetry.** The vertical line dividing the parabola into two equal portions is called the **line of symmetry.** All parabolas have a **vertex,** the ordered pair that represents the bottom (or the top) of the curve.

The vertex of a parabola is the ordered pair \begin{align*}(h, k)\end{align*}

Because the line of symmetry is a vertical line, its equation has the form \begin{align*}x=h\end{align*}*the* \begin{align*}x-\end{align*}*coordinate of the vertex.*

As with linear equations, the \begin{align*}x\end{align*}

An equation of the form \begin{align*}y=ax^2\end{align*} forms a parabola.

If \begin{align*}a\end{align*} is positive, the parabola will open **upward.** The vertex will be a **minimum.**

If \begin{align*}a\end{align*} is negative, the parabola will open **downward.** The vertex will be a **maximum.**

The variable \begin{align*}a\end{align*} in the equation above is called the **leading coefficient** of the quadratic equation. Not only will it tell you if the parabola opens up or down, but it will also tell you the width.

If \begin{align*}a>1\end{align*} or \begin{align*}a<-1\end{align*}, the parabola will be **narrow** about the line of symmetry.

If \begin{align*}-1<a<1\end{align*}, the parabola will be **wide** about the line of symmetry.

#### Example B

*Find the \begin{align*}x\end{align*}-intercepts of the quadratic function \begin{align*}y=x^2+5x-6\end{align*}.*

**Solution:**

To find the \begin{align*}x\end{align*}-intercepts, set \begin{align*}y\end{align*} equal to zero and solve for \begin{align*}x\end{align*} by factoring.

\begin{align*}0=x^2+5x-6=x^2+-1x+6x-6=x(x-1)+6(x-1)=(x+6)(x-1)\end{align*}

This means that

\begin{align*}0=x+6 \Rightarrow x=-6\end{align*} or \begin{align*}0=x-1 \Rightarrow x=1\end{align*}.

Thus the \begin{align*}x\end{align*}-intercepts are -6 and 1.

**Finding the Vertex of a Quadratic Equation in Standard Form**

The \begin{align*}x-\end{align*}coordinate of the vertex of \begin{align*}0=ax^2+bx+c\end{align*} is \begin{align*}x=-\frac{b}{2a}\end{align*}.

#### Example C

*Determine the direction, shape and vertex of the parabola formed by* \begin{align*}y=-\frac{1}{2} x^2\end{align*}.

**Solution:**

The value of \begin{align*}a\end{align*} in the quadratic equation is \begin{align*}-\frac{1}{2}\end{align*}.

- Because \begin{align*}a\end{align*} is negative, the parabola opens downward.
- Because \begin{align*}a\end{align*} is between –1 and 1, the parabola is wide about its line of symmetry.
- Because there is no \begin{align*}b\end{align*} term, \begin{align*}b=0\end{align*}. Substituting this into the equation for the \begin{align*}x\end{align*}-coordinate of the vertex, \begin{align*}x=-\frac{b}{2a}=-\frac{0}{2a}=0\end{align*}. (Note: It does not matter what \begin{align*}a\end{align*} equals; since \begin{align*}b=0\end{align*}, the fraction equals zero.) To find the \begin{align*}y\end{align*}-coordinate, substitute the \begin{align*}x\end{align*}-coordinate into the equation:

\begin{align*}y=-\frac{1}{2} x^2=-\frac{1}{2}\cdot 0^2=0\end{align*}

The vertex is \begin{align*}(0,0)\end{align*}.

**Domain and Range**

Several times throughout this textbook, you have experienced the terms **domain** and **range.** Remember:

- Domain is the set of all inputs (\begin{align*}x-\end{align*}coordinates).
- Range is the set of all outputs (\begin{align*}y-\end{align*}coordinates).

The domain of every quadratic equation is all real numbers \begin{align*}(\mathbb{R})\end{align*}. The range of a parabola depends upon whether the parabola opens up or down.

If \begin{align*}a\end{align*} is positive, the range will be \begin{align*}y \ge k\end{align*}.

If \begin{align*}a\end{align*} is negative, the range will be \begin{align*}y \le k\end{align*}, where \begin{align*}k= \text{the }y-\end{align*}*coordinate of the vertex.*

#### Example D

*Find the range of the quadratic function \begin{align*}y=-2x^2+16x+5\end{align*}.*

**Solution:**

To find the range, we must find the \begin{align*}y\end{align*}-value of the vertex. Using the formula given above, we can find the \begin{align*}x\end{align*}-value of the vertex, and use that to find the \begin{align*}y\end{align*}-value of the vertex.

Since the \begin{align*}x\end{align*}-value of the vertex is \begin{align*}x=-\frac{b}{2a}\end{align*}, we get \begin{align*}x=-\frac{16}{2\cdot -2}=4.\end{align*}

Now, substitute 4 into the function:

\begin{align*}y=-2x^2+6x+5=-2(4)^2+6(4)+5=-2(16)+24+5=-32+29=-3\end{align*}

The \begin{align*}y\end{align*}-value of the vertex is -3, and since \begin{align*}a=-2\end{align*}, the parabola is facing down, so -3 is the highest possible value for the range.

Thus, the range is \begin{align*}y\le -3\end{align*}.

### Guided Practice

*Determine the direction, vertex and range of \begin{align*}y=7x^2+14x-9\end{align*}.*

**Solution:**

Since \begin{align*}a=7\end{align*} is positive, the direction of the parabola is upward. Now we find the vertex:

\begin{align*}x=-\frac{b}{2a}=-\frac{14}{2\cdot 7}=-\frac{14}{14}=-1.\end{align*}

Now, substitute \begin{align*}x=-1\end{align*} into the quadratic function:

\begin{align*}y=7x^2+14x-9=7(-1)^2+14(-1)-9=7(1)-14-9=7-23=-16.\end{align*}

Thus, the vertex is (-1, -16).

Since the parabola faces up, and the \begin{align*}y\end{align*}-value of the vertex is -16, the range is \begin{align*}y\ge -16\end{align*}.

### Practice

Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Graphs of Quadratic Functions (16:05)

- Define the following terms in your own words.
- Vertex
- Line of symmetry
- Parabola
- Minimum
- Maximum

- Without graphing, how can you tell if \begin{align*}y=ax^2+bx+c\end{align*} opens up or down?

Graph the following equations by making a table. Let \begin{align*}-3 \le x \le 3\end{align*}. Determine the range of each equation.

- \begin{align*}y=2x^2\end{align*}
- \begin{align*}y=-x^2\end{align*}
- \begin{align*}y=x^2-2x+3\end{align*}
- \begin{align*}y=2x^2+4x+1\end{align*}
- \begin{align*}y=-x^2+3\end{align*}
- \begin{align*}y=x^2-8x+3\end{align*}
- \begin{align*}y=x^2-4\end{align*}

Does the graph of the parabola open up or down?

- \begin{align*}y=-2x^2-2x-3\end{align*}
- \begin{align*}y=3x^2\end{align*}
- \begin{align*}y=16-4x^2\end{align*}

Find the \begin{align*}x-\end{align*}coordinate of the vertex of the following equations.

- \begin{align*}x^2-14x+45=0\end{align*}
- \begin{align*}8x^2-16x-42=0\end{align*}
- \begin{align*}4x^2+16x+12=0\end{align*}
- \begin{align*}x^2+2x-15=0\end{align*}

Graph the following functions by making a table of values. Use the vertex and \begin{align*}x-\end{align*}intercepts to help you pick values for the table.

- \begin{align*}y=4x^2-4\end{align*}
- \begin{align*}y=-x^2+x+12\end{align*}
- \begin{align*}y=2x^2+10x+8\end{align*}
- \begin{align*}y=\frac{1}{2} x^2-2x\end{align*}
- \begin{align*}y=x-2x^2\end{align*}
- \begin{align*}y=4x^2-8x+4\end{align*}