Look at the graph below. Does the graph represent a function? Do you know the name of the graph? Do you know what makes the green point special? Do you notice any symmetry in the graph? Can you state the domain and range for the relation?

### Watch This

Khan Academy Quadratic Functions 1

### Guidance

Until now you have been dealing with linear functions. The highest exponent of the independent variable (\begin{align*}x\end{align*}) has been one and the graphs have been straight lines. Here you will be learning about quadratic functions. A quadratic function is one of the form @$\begin{align*}y=ax^2+bx+c\end{align*}@$ where @$\begin{align*}a, b\end{align*}@$ and @$\begin{align*}c\end{align*}@$ are real numbers and @$\begin{align*}a \ne 0\end{align*}@$. The highest exponent of the independent variable is two. When graphed, a quadratic function creates a parabola that looks like this: or like this:

You can create your own graph by plotting the points created from a table of values. The most basic quadratic function is @$\begin{align*}y=x^2\end{align*}@$. The easiest way to make a table for this function is to use the domain @$\begin{align*}\{x|-3 \le x \le 3, x \ \in \ Z\}\end{align*}@$ for the table.

A parabola has a turning point known as the vertex. The vertex is the minimum value of the parabola if it opens upward and the maximum value if the parabola opens downward. When the graph opens downward, the @$\begin{align*}y\end{align*}@$-values in the base table change to negative values. The basic quadratic function that opens downward has the equation @$\begin{align*}y=-x^2\end{align*}@$.

All parabolas have an axis of symmetry. The axis of symmetry is the vertical line that passes through the vertex of the parabola. The equation for the axis of symmetry is always @$\begin{align*}x=\alpha\end{align*}@$, where @$\begin{align*}\alpha\end{align*}@$ is the @$\begin{align*}x-\end{align*}@$coordinate of the vertex.

#### Example A

For the basic quadratic function @$\begin{align*}y=x^2\end{align*}@$, complete a table such that @$\begin{align*}\{x|-3 \le x \le 3, x \ \in \ Z\}\end{align*}@$.

**Solution:**

To complete the table of values, substitute the given @$\begin{align*}x\end{align*}@$-values into the function @$\begin{align*}y=x^2\end{align*}@$. If you are using a calculator, insert all numbers, especially negative numbers, inside parentheses before squaring them. The operation that needs to be done is @$\begin{align*}(-3)(-3)\end{align*}@$ NOT @$\begin{align*}-(3)(3)\end{align*}@$.

@$$\begin{align*}y&=x^2 && y=x^2 && y=x^2 && y=x^2\\ y&=(-3)^2 && y=(-2)^2 && y=(-1)^2 && y=(0)^2\\ y&={\color{red}9} && y={\color{red}4} && y={\color{red}1} && y={\color{red}0}\\ \\ y&=x^2 && y=x^2 && y=x^2\\ y&=(1)^2 && y=(2)^2 && y=(3)^2\\ y&={\color{red}1} && y={\color{red}4} && y={\color{red}9}\\\end{align*}@$$

@$\begin{align*}X\end{align*}@$ |
@$\begin{align*}Y\end{align*}@$ |
---|---|

@$\begin{align*}-3\end{align*}@$ | @$\begin{align*}{\color{red}9}\end{align*}@$ |

@$\begin{align*}-2\end{align*}@$ | @$\begin{align*}{\color{red}4}\end{align*}@$ |

@$\begin{align*}-1\end{align*}@$ | @$\begin{align*}{\color{red}1}\end{align*}@$ |

@$\begin{align*}0\end{align*}@$ | @$\begin{align*}{\color{red}0}\end{align*}@$ |

@$\begin{align*}1\end{align*}@$ | @$\begin{align*}{\color{red}1}\end{align*}@$ |

@$\begin{align*}2\end{align*}@$ | @$\begin{align*}{\color{red}4}\end{align*}@$ |

@$\begin{align*}3\end{align*}@$ | @$\begin{align*}{\color{red}9}\end{align*}@$ |

#### Example B

On a Cartesian plane, plot the points from the table for @$\begin{align*}y=x^2\end{align*}@$.

**Solution:**

The plotted points cannot be joined to form a continuous curve. To join the points, begin with the point (–3, 9) or the point (3, 9) and without lifting your pencil, draw a smooth curve. The image should look like the following graph.

The arrows indicate the direction of the pencil as the points are joined. If the pencil is not moved off the paper, the temptation to join the points with a series of straight lines will be decreased. The points must be joined with a smooth curve that does not extend below the lowest point of the graph. In the above graph, the curve cannot go below the point (0, 0).

#### Example C

What are some unique characteristics of the graph of @$\begin{align*}y=x^2\end{align*}@$?

**Solution:**

- The green point is located at the lowest point on the image. The curve does not go below this point.
- Every red point on the left side of the image has a corresponding blue point on the right side of the image.
- If the image was folded left to right along the @$\begin{align*}y\end{align*}@$-axis that passes through the green point, each red point would land on each corresponding blue point.
- The sides of the image extend upward.
- The red and the blue points are plotted to the right and to the left of the green point. The points are plotted left and right one and up one; left and right two and up four, left and right 3 and up nine.

#### Concept Problem Revisited

The green point is the lowest point on the curve. The smooth curve is called a *parabola* and it is the image produced when the basic quadratic function is plotted on a Cartesian grid. The green point is known as the *vertex* of the parabola. The *vertex* is the turning point of the graph.

For the graph of @$\begin{align*}y=x^2\end{align*}@$, the vertex is (0, 0) and the parabola has a minimum value of zero which is indicated by the @$\begin{align*}y\end{align*}@$-value of the vertex. The parabola opens upward since the @$\begin{align*}y\end{align*}@$-values in the table of values are 0, 1, 4 and 9. The @$\begin{align*}y\end{align*}@$-axis for this graph is actually the axis of symmetry. The axis of symmetry is the vertical line that passes through the vertex of the parabola. The parabola is symmetrical about this line. The equation for this axis of symmetry is @$\begin{align*}x = 0\end{align*}@$. If the parabola were to open downward, the vertex would be the highest point of the graph. Therefore the image would have a maximum value of zero.

The domain for all parabolas is @$\begin{align*}D=\{x|x \ \in \ R\}\end{align*}@$. The range for the above parabola is @$\begin{align*}R=\{y|y \ge 0, y \ \in \ R\}\end{align*}@$.

### Guided Practice

1. If the graph of @$\begin{align*}y=x^2\end{align*}@$ opened downward, what changes would exist in the base table of values?

2. If the graph of @$\begin{align*}y=x^2\end{align*}@$ opened downward, what changes would exist in the basic quadratic function?

3. Draw the image of the basic quadratic function that opens downward. State the domain and range for this parabola.

**Answers:**

1. If the parabola were to open downward, the @$\begin{align*}x\end{align*}@$-values would not change. The @$\begin{align*}y\end{align*}@$-values would become negative values. The points would be plotted from the vertex as: right and left one and down one; right and left two and down four; right and left three and down nine. The table of values would be

@$\begin{align*}X\end{align*}@$ |
@$\begin{align*}Y\end{align*}@$ |
---|---|

@$\begin{align*}-3\end{align*}@$ | @$\begin{align*}{\color{red}-9}\end{align*}@$ |

@$\begin{align*}-2\end{align*}@$ | @$\begin{align*}{\color{red}-4}\end{align*}@$ |

@$\begin{align*}-1\end{align*}@$ | @$\begin{align*}{\color{red}-1}\end{align*}@$ |

@$\begin{align*}0\end{align*}@$ | @$\begin{align*}{\color{red}0}\end{align*}@$ |

@$\begin{align*}1\end{align*}@$ | @$\begin{align*}{\color{red}-1}\end{align*}@$ |

@$\begin{align*}2\end{align*}@$ | @$\begin{align*}{\color{red}-4}\end{align*}@$ |

@$\begin{align*}3\end{align*}@$ | @$\begin{align*}{\color{red}-9}\end{align*}@$ |

2. To match the table of values, the basic quadratic function would have to be written as @$\begin{align*}y=-x^2.\end{align*}@$

3.

The domain is @$\begin{align*}D=\{x|x \ \in \ R\}\end{align*}@$. The range for this parabola is @$\begin{align*}R=\{y|y \le 0, y \ \in \ R\}\end{align*}@$.

### Explore More

Complete the following statements in the space provided.

- The name given to the graph of @$\begin{align*}y=x^2\end{align*}@$ is ____________________.
- The domain of the graph of @$\begin{align*}y=x^2\end{align*}@$ is ____________________.
- If the vertex of a parabola was (–3, 5), the equation of the axis of symmetry would be ____________________.
- A parabola has a maximum value when it opens ____________________.
- The point (–2, 4) on the graph of @$\begin{align*}y=x^2\end{align*}@$ has a corresponding point at ____________________.
- The range of the graph of @$\begin{align*}y=-x^2\end{align*}@$ is ____________________.
- If the table of values for the basic quadratic function included 4 and –4 as @$\begin{align*}x\end{align*}@$-values, the @$\begin{align*}y\end{align*}@$-value(s) would be ____________________.
- The vertical line that passes through the vertex of a parabola is called ____________________.
- A minimum value exists when a parabola opens ____________________.
- The turning point of the graph of @$\begin{align*}y=x^2\end{align*}@$ is called the ____________________.

- Make a sketch of the function @$\begin{align*}y=x^2-2x+1\end{align*}@$ by first making a table and then plotting the points. What is the vertex of this parabola?
- Make a sketch of the function @$\begin{align*}y=x^2+2x-3\end{align*}@$ by first making a table and then plotting the points. What is the vertex of this parabola?
- Make a sketch of the function @$\begin{align*}y=x^2-4x+5\end{align*}@$ by first making a table and then plotting the points. What is the vertex of this parabola?
- Make a sketch of the function @$\begin{align*}y=-x^2+2x+1\end{align*}@$ by first making a table and then plotting the points. What is the vertex of this parabola?
- Make a sketch of the function @$\begin{align*}y=-x^2-4x\end{align*}@$ by first making a table and then plotting the points. What is the vertex of this parabola?