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{{ printedBook.courseTrack.name }} {{ printedBook.name }} There are several ways to write the rule of a quadratic function. Each form highlights certain characteristics of the parabola. Standard form is expressed as follows.

$y=ax^2+bx+c$

When a quadratic function is written in standard form, it's possible to use $a,$ $b,$ and $c$ to determine characteristics of its graph. $\begin{aligned} \textbf{direction} &: \text{upward when } a>0, \\ &\phantom{:} \text{downward when } a<0 \\ \mathbf{y} \textbf{-intercept} &: (0,c) \\ \textbf{axis of symmetry} &: x=\text{-}\dfrac{b}{2a} \end{aligned}$

The direction of the graph is determined by the sign of $a.$ To understand why, consider the quadratic function $y=ax^2.$ Since all squares are positive, $x^2$ will always be positive. When $a$ is positive, then $ax^2$ is also positive. Thus, when moving away from the origin in either direction, the graph extends upward. Similarly, when $a$ is negative, $ax^2$ will be negative. Thus, the graph will extend downward for all $x$-values.

The $y$-intercept of a quadratic function is given by $c,$ specifically at $(0,c).$ This is because substituting $x=0$ into standard form yields the following. $\begin{aligned} y&=ax^2+bx+c\\ y&=a \cdot {\color{#0000FF}{0}}^2 + b \cdot {\color{#0000FF}{0}} + c \\ y&= 0+0+c \\ y&=c \end{aligned}$

For all quadratic functions, the axis of symmetry will always intersect the parabola at its vertex. Additionally, two points with the same $y$-coordinate will always be equidistant from the axis of symmetry. Move the three points to see how a parabola that passes through them looks.

Draw the graph

The function $f(x)=3x-0.5x^2$ describes the height of the mouth of a tunnel. Here, $x$ is the distance from the lower left corner, and both $x$ and $f(x)$ are in meters. Complete the table of values to graph the function and determine the width and height of the tunnel.

$x$ | $y$ |
---|---|

$0$ | |

$1$ | |

$2$ | |

$3$ | |

$4$ | |

$5$ | |

$6$ |

Show Solution

To begin, we can complete the table of values by substituting the given $x$-values into the function rule. We'll start with $x=0.$
We can perform the same process for the other $x$-values.

$x=3$ means that the point on the parabola whose $x$-coordinate is $3$ is the maximum. In the table, we found that when $x=3, y=4.5.$ Thus, the height of the tunnel is $4.5$ meters.

$f(x)=3x-0.5x^2$

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

$f({\color{#0000FF}{0}})=3\cdot {\color{#0000FF}{0}}-0.5\cdot {\color{#0000FF}{0}}^2$

CalcPowProdCalculate power and product

$f(0)=0$

$x$ | $3x-0.5x^2$ | $f(x)$ |
---|---|---|

${\color{#0000FF}{1}}$ | $3\cdot {\color{#0000FF}{1}}-0.5\cdot {\color{#0000FF}{1}}^2$ | $2.5$ |

${\color{#0000FF}{2}}$ | $3\cdot {\color{#0000FF}{2}}-0.5\cdot {\color{#0000FF}{2}}^2$ | $4$ |

${\color{#0000FF}{3}}$ | $3\cdot {\color{#0000FF}{3}}-0.5\cdot {\color{#0000FF}{3}}^2$ | $4.5$ |

${\color{#0000FF}{4}}$ | $3\cdot {\color{#0000FF}{4}}-0.5\cdot {\color{#0000FF}{4}}^2$ | $4$ |

${\color{#0000FF}{5}}$ | $3\cdot {\color{#0000FF}{5}}-0.5\cdot {\color{#0000FF}{5}}^2$ | $2.5$ |

${\color{#0000FF}{6}}$ | $3\cdot {\color{#0000FF}{5}}-0.5\cdot {\color{#0000FF}{6}}^2$ | $0$ |

To graph the function, we can plot the points and connect them with a smooth curve.

The graph gives an approximation of the height and width of the tunnel. We can think of the $x$-axis as the ground. Thus, the distance between the $x$-axis and the vertex, which is a maximum, gives the height of the tunnel, and the distance between the zeros gives the width of the tunnel.

From this graph, we can see that the height appears to be $4.5$ meters and that the width appears to be $6$ units. Because the zeros were found algebraically to be $(0,0) \quad \text{and} \quad (6,0),$ we can definitively state that the width of the tunnel is $6$ meters. What remains is to algebraically determine the coordinate of the maximum (vertex). The $x$-coordinate of the vertex can be found using the formula for the axis of symmetry, because the axis of symmetry intersects the parabola at its vertex. $x=\text{-} \frac{b}{2a}$ Rearranging the function rule so it is shown in standard form gives $f(x)=3x-0.5x^2 \quad \leftrightarrow \quad f(x)=\text{-} 0.5x^2+3x.$ Thus, $a=\text{-} 0.5$ and $b=3.$ These values can be substituted into the formula.$x=\text{-}\dfrac{b}{2a}$

$x=\text{-}\dfrac{{\color{#009600}{3}}}{2 ({\color{#0000FF}{\text{-} 0.5}})}$

MultiplyMultiply

$x=\text{-}\dfrac{3}{\text{-} 1}$

RemoveNegFracAndDenom$\text{-} \dfrac{a}{\text{-} b}= \dfrac{a}{b}$

$x=\dfrac{3}{1}$

DivByOne$\dfrac{a}{1}=a$

$x=3$

It is possible to graph a quadratic function in standard form by using the characteristics given by its function rule. Consider the function
$f(x)=x^2-4x+3.$
### 1

The axis of symmetry can be found using $a$ and $b.$ Here, $a=1$ and $b=\text{-}4.$
The axis of symmetry for this function is $x=2.$
### 2

The axis of symmetry always intersects the parabola at its vertex. Thus, $x=2$ is the $x$-coordinate of the vertex. Use this to determine the corresponding $y$-coordinate.
The vertex of the function is $(2,\text{-}1).$
### 3

The $y$-intercept can be found using $c.$ In this case $c=3,$ meaning that the $y$-intercept is $(0,3).$ Add this point to the graph.

### 4

Since the axis of symmetry divides the graph into two mirror images, there exists another point on the other side of the axis of symmetry with the same $y$-value as the $y$-intercept. These points are equidistant from the axis of symmetry.

### 5

Identify and graph the axis of symmetry

$x=\text{-}\dfrac{b}{2a}$

$x=\text{-}\dfrac{{\color{#009600}{\text{-}4}}}{2\cdot{\color{#0000FF}{1}}}$

MultiplyMultiply

$x=\text{-}\dfrac{\text{-}4}{2}$

CalcQuotCalculate quotient

$x=\text{-}(\text{-}2)$

NegNeg$\text{-} (\text{-} a)=a$

$x=2$

Determine and plot the vertex

$f(x)=x^2-4x+3$

Substitute$x={\color{#0000FF}{2}}$

$f({\color{#0000FF}{2}})={\color{#0000FF}{2}}^2-4\cdot{\color{#0000FF}{2}}+3$

CalcPowProdCalculate power and product

$f(2)=4-8+3$

AddSubTermsAdd and subtract terms

$f(2)=\text{-}1$

Determine and plot the $y$-intercept

Reflect the $y$-intercept across the axis of symmetry

Draw the parabola

Now, the general shape of the parabola can be seen. Connect the points with a smooth curve.

If the dependent variable of a quadratic function is exchanged for a constant, say $D,$ the result is a quadratic equation: $ax^2+bx+c=D.$

This type of equation can be solved graphically. This is done by first plotting the function $y = ax^2+bx+c,$ then finding the $x$-coordinate of the point(s) on the graph that has the $y$-coordinate $D.$ The $x$-coordinate(s) is the solution to the equation. {{ 'mldesktop-placeholder-grade' | message }} {{ article.displayTitle }}!

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