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Quadratic function rules can be expressed in different ways. Each form is useful because it highlights different characteristics of the parabola. As the name implies, vertex form gives the vertex of the parabola.

**Vertex form** is an algebraic format used to express quadratic function rules.

$y=a(x−h)_{2}+k$

In this form, $a$ gives the direction of the parabola. When $a>0,$ the parabola faces upward and when $a<0,$ it faces downward. The vertex of the parabola lies at $(h,k),$ and the axis of symmetry is $x=h.$ Consider the graph of $f(x)=-21 (x+4)_{2}+8.$

From the graph, we can connect the following characterisitcs to the function rule. $directionvertexaxis of symmetry :downward:(-4,8):x=-4 →a<0→h=-4,k=8→h=-4 $

Notice that although the factor in the function rule shows $(x+4)_{2},$ $h$ is actually equal to $-4.$ This coincides with a horizontal translation of a quadratic function.Writing quadratic functions in vertex form, $y=a(x−h)_{2}+k,$ is advantageous because it clearly presents three characteristics of the function. $directionvertexaxis of symmetry :a>0⇒upward,a<0⇒downward:(h,k):x=h $ It is possible to graph a quadratic function using its characteristics. Consider the function $y=(x−2)_{2}−4.$

Identify and plot the vertex

Draw the axis of symmetry

Determine and plot the $y$-intercept

$y=(x−2)_{2}−4$

Substitute $x$ for $0$ and evaluate

Substitute$x=0$

$y=(0−2)_{2}−4$

SubTermSubtract term

$y=(-2)_{2}−4$

CalcPowCalculate power

$y=4−4$

SubTermSubtract term

$y=0$

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

The axis of symmetry divides the parabola into two mirror images. Thus, points on one side of the parabola can be reflected across the axis of symmetry. Notice that the $y$-intercept is $2$ units away from the axis of symmetry.

There exists another point directly across from the $y$-intercept that is also $2$ units from the axis of symmetry.

Draw the parabola

The characteristics of two functions, $f(x)$ and $g(x),$ are described below. Determine which of the function rules could represent $g$ and $h.$ $-Axis of symmetryVertex f(x).x=1(1,3) g(x)-x=-1(-1,-3) $

Number | Function rule |
---|---|

$1$ | $y=-(x−1)_{2}+3$ |

$2$ | $y=(x−1)_{2}−3$ |

$3$ | $y=-(x−1)_{2}−3$ |

$4$ | $y=(x−1)_{2}+3$ |

$5$ | $y=(x+1)_{2}+3$ |

$6$ | $y=(x+1)_{2}−3$ |

$7$ | $y=-(x+1)_{2}+3$ |

$8$ | $y=-(x+1)_{2}−3$ |

Show Solution

When a quadratic function is given in vertex form, $y=a(x−h)_{2}+k$, both the vertex and the axis of symmetry can be easily seen. $vertexaxis of symmetry :(h,k):x=h $ It is given that the axis of symmetry for $g(x)$ is $x=1$ and for $f(x)$ it is $x=-1.$ Since the axis of symmetry is given as $x=h,$ this gives $h=1$ for $g$ and $h=-1$ for $f.$ Thus, we can begin to write the function rules for $g(x)$ and $f(x)$ in vertex form as follows. Note that $(x−(-1))$ becomes $(x+1).$ $g(x)=a(x+1)_{2}+kf(x)=a(x−1)_{2}+k $ We can use the given vertices to determine the values of $k.$ The vertex of $g$ is $(-1,-3).$ This gives $k=-3.$ Similarly, $f(x)$'s vertex $(1,3)$ gives $k=3.$ Thus, we can write the rules as $f(x)g(x) =a(x+1)_{2}+3=a(x−1)_{2}−3. $ Comparing these rules to the table above, it can be seen that rule number $2$ and $3$ can represent $g(x)$ and $5$ and $7$ can represent $f(x).$

Number | Function rule |
---|---|

$2$ | $y=(x−1)_{2}−3$ |

$3$ | $y=-(x−1)_{2}−3$ |

$5$ | $y=(x+1)_{2}+3$ |

$7$ | $y=-(x+1)_{2}+3$ |

There are two candidates remaining for each function. The rules with $a=-1$ correspond to a downward parabola, whereas the rules with $a=1$ correspond to an upward parabola. To determine if any option is completely correct, we'd need more information about the parabola, specifically its direction or if its vertex is a minimum or a maximum.

Write the rule for the quadratic function that has the vertex $(5,-1)$ and passes through the point $(10,-11)$.

Show Solution

It's possible to write the rule of a quadratic function in vertex form $(x)=a(x−h)_{2}+k$ using two points. Let's start with the vertex. In vertex form, $(h,k)$ is the coordinate of the vertex. Knowing this, we can substitute $h=5$ and $k=-1$ into the rule.
$f(x)=a(x−5)_{2}−1$
Notice that the rule is incomplete. We can use the other point to determine the value of $a.$ Since $(10,-11)$ lies on the parabola, $x=10$ and $y=-11$ can be substituted into the rule. Then, we can solve for $a.$
Thus, we can write the complete function rule as $f(x)=-0.4(x−5)_{2}−1.$
To verify that $f$ passes through $(5,1)$ and $(10,-11),$ we can graph the parabola and mark the points.

$y(x)=a(x−5)_{2}−1$

$-11=a(10−5)_{2}−1$

SubTermSubtract term

$-11=a(5)_{2}−1$

CalcPowCalculate power

$-11=25a−1$

AddEqn$LHS+1=RHS+1$

$-10=25a$

DivEqn$LHS/25=RHS/25$

$25-10 =a$

SimpQuotSimplify quotient

$-0.4=a$

RearrangeEqnRearrange equation

$a=-0.4$

Since the parabola passes through the given points, the created rule is correct.

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