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Like linear and exponential functions, quadratic functions are a unique type of functions that have specific qualities in common. Analyzing these functions in terms of their *characteristics* allows important information to be learned.

A quadratic function is a function of degree $2.$ That means that the highest exponent of the independent variable is $2.$ The simplest quadratic function is $y=x^2,$ and the graph of any quadratic function is a parabola.

The inherent shape of parabolas gives rise to several characteristics that all quadratic functions have in common.

A parabola either opens **upward** or **downward**. This is called its direction.

Because a parabola either opens upward or downward, there is always one point that is the absolute maximum or absolute minimum of the function. This point is called the vertex.

At the vertex, the function changes from increasing to decreasing, or vice versa.

All parabolas are symmetric, meaning there exists a line that divides the graph into two mirror images. For quadratic functions, that line is always parallel to the $y$-axis, and is called the axis of symmetry.

The axis of symmetry always intersects the vertex of the parabola, and is written as a vertical line, where $h$ can be any real number.

$x=h$

Depending on its rule, a parabola can intersect the $x$-axis at $0,$ $1,$ or $2$ points. Since the function's value at an $x$-intercept is always $0,$ these points are called zeros, or sometimes roots.

For the quadratic function $y=x^2-4x,$ create a table of values to graph it. Then determine its direction, vertex, zeros, and axis of symmetry.

Show Solution

To begin, we'll use the function rule to create a table of values. Then, to graph the function, we'll plot the points from the table. Let's start with $x=0.$
Thus, the point $(0,0)$ lies on the parabola. We can perform the same calculations for other $x$-values around $x=0.$

$y=x^2-4x$

Substitute$x={\color{#0000FF}{0}}$

$y={\color{#0000FF}{0}}^2-4({\color{#0000FF}{0}})$

CalcPowProdCalculate power and product

$y=0$

$x$ | $x^2-4x$ | $y$ |
---|---|---|

${\color{#0000FF}{\text{-} 2}}$ | $({\color{#0000FF}{\text{-} 2}})^2-4({\color{#0000FF}{\text{-} 2}})$ | $12$ |

${\color{#0000FF}{\text{-}1}}$ | $({\color{#0000FF}{\text{-} 1}})^2-4({\color{#0000FF}{\text{-} 1}})$ | $5$ |

${\color{#0000FF}{0}}$ | ${\color{#0000FF}{0}}^2-4({\color{#0000FF}{0}})$ | $0$ |

${\color{#0000FF}{1}}$ | ${\color{#0000FF}{1}}^2-4( {\color{#0000FF}{1}})$ | $\text{-} 3$ |

${\color{#0000FF}{2}}$ | ${\color{#0000FF}{2}}^2-4( {\color{#0000FF}{2}})$ | $\text{-} 4$ |

We'll plot these points on a coordinate plane.

We can start to see the left-hand side of the parabola. Let's add a few more $x$-values to the table to determine a more complete shape.

$x$ | $x^2-4x$ | $y$ |
---|---|---|

${\color{#0000FF}{3}}$ | ${\color{#0000FF}{3}}^2-4 ({\color{#0000FF}{3}})$ | $\text{-} 3$ |

${\color{#0000FF}{4}}$ | ${\color{#0000FF}{4}}^2-4 ({\color{#0000FF}{4}})$ | $0$ |

${\color{#0000FF}{5}}$ | ${\color{#0000FF}{5}}^2-4 ({\color{#0000FF}{5}})$ | $5$ |

We'll add these points to the coordinate system as well.

Looking at the points, we now see both sides of the parabola. We can connect the points with a smooth curve.

The graph can be used to describe the desired characteristics of the parabola.

$\begin{aligned} \textbf{direction} &: \text{upward} \\ \textbf{axis of symmetry} &: x=2\\ \textbf{vertex} &: (2,\text{-} 4)\\ \textbf{zeros} &: (0,0) \text{ and } (4,0) \end{aligned}$

Three quadratic functions are graphed in the coordinate plane.

For each graph, match it with the corresponding characteristics. $\begin{aligned} \textbf{direction:}& \quad \text{upward,} \quad \text{downward} \\ \textbf{vertex:}& \quad \text{minimum,} \quad \text{maximum} \\ \textbf{vertex:}& \quad (\text{-}2,2), \quad (0,\text{-}6), \quad (2,\text{-}4) \\ \textbf{axis of symmetry:}& \quad x=\text{-}2, \quad x=2, \quad x=0 \\ \textbf{$y$-intercept:}& \quad y=\text{-}6, \quad y=\text{-}2, \ \ \ y=0, \quad y=4\\ \textbf{zero:}& \quad x=0, \ \quad x=4, \quad x=\text{-}6 \end{aligned}$

Show Solution

Instead of looking at each function separately, we'll look at the characteristics individually and summarize our findings in a table at the end.

First, let's consider the direction of the parabolas. We can see that $A$ and $C$ open upward, and that $B$ opens downward. The direction of a parabola determines whether the vertex is a minimum or a maximum. Thus, the vertices of $A$ and $C$ are minimums while the vertex of $B$ is a maximum.

The vertex for each graph is found at the minimum or maximum of the function. For $A,$ the vertex lies at $(2,\text{-} 4).$ Similarly, $B$'s vertex lies at $(\text{-} 2,2)$, and $C$ has its vertex at $(0,\text{-}6).$

The axis of symmetry is the vertical line that intersects the vertex. Therefore, the axis of symmetry for graph $A$ is $x=2,$ for $B$ it's $x=\text{-}2$ and for $C$ it is $x=0.$

The $y$-intercepts are found where the parabolas intercept the $y$-axis. $A$ has the $y$-intercept $y=0$, $B$ has $y=\text{-} 2$ and $C$ has $y=\text{-} 6$. One of the options, $y=4,$ does not coincide with any graph.

The zeros are found where the parabolas intercept the $x$-axis. Then, function $A$ has zeros $x=4$ and $x=0$, which are two of the zeros given in the prompt. Neither function $B$ nor function $C$ has a zero at $x=\text{-} 6$.

We summarize what we have learned about the characteristics of the three quadratic functions.

Function | $A$ | $B$ | $C$ |
---|---|---|---|

direction | upward | downward | upward |

max/min | minimum | maximum | minimum |

vertex | $(2,\text{-} 4)$ | $(\text{-}2,2)$ | $(0,\text{-} 6)$ |

axis of symmetry | $x=2$ | $x=\text{-}2$ | $x=0$ |

$y$-intercept | $y=0$ | $y=\text{-}2$ | $y=\text{-}6$ |

zeros | $x=0$ and $x=4$ | not applicable | not applicable |

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