Vertex of a parabola

Quadratic equations: Drawing parabolas

Vertex of a parabola

A quadratic with a parabola opening downwards has a highest, maximum point. A quadratic with a parabola opening upwards has a lowest, minimum point. This highest and lowest point is also called the vertex of the graph. We will take a look at two different methods to determine the vertex.

Determining the vertex by completing the square

The vertex of a quadratic \[y=a\left(x-\blue p\right)^2+\green q\] is the point \[\rv{\blue p,\green q}\]

To determine the vertex of a quadratic in a different form, we can complete the square in such way so that we can rewrite the formula in the form \[y=a\left(x-\blue p\right)^2+\green q\]

geogebra plaatje

Coefficent for quadratic termsEven if the coefficient of the term #x^2# is not equal to #1#, we can complete the square.

	Procedure	Example
	We complete the square in the expression #\blue ax^2+\green bx+\purple c#.	#\blue4x^2+\green{24}x+\purple9#
Step 1	Rewrite the expression to #\blue a\left(x+\tfrac{\green b}{\blue a}x\right)+\purple c#.	#=\blue 4\left(x^2+6x\right)+\purple 9#
Step 2	Complete the square of #\left(x+\tfrac{\green b}{\blue a}x\right)#.	#=\blue 4 \left(\left(x+3\right)^2-9\right)+\purple 9#
Step 3	Expand the outer brackets.	#=\blue 4\left(x+3\right)^2-36+\purple 9#
Step 4	Add the constants.	#=\blue 4\left(x+3\right)^2-27#

Example

Take a look a the quadratic #y=-3x^2+6x+2#.

To determine the vertex, we rewrite the formula in the form #a\left(x-\blue p\right)^2+\green q#. This is done like this:

\[\begin{array}{rcl}y&=&-3x^2+6x+2 \\ &&\phantom{xxxxx}\blue{\text{the given formula }} \\&=&-3 \left(x^2-2x\right)+2 \\ &&\phantom{xxxxx}\blue{-3 \text{ expanded the brackets }}\\&=&-3 \left(\left(x-1\right)^2-1\right)+2 \\ &&\phantom{xxxxx}\blue{\text{completed the square}}\\ &=&-3\left(x-1\right)^2+3+2 \\ &&\phantom{xxxxx}\blue{\text{brackets expanded}} \\ &=&-3\left(x-1\right)^2+5 \\ &&\phantom{xxxxx}\blue{\text{simplified}} \end{array}\]

Hence, we can write #y=-3x^2+6x+2# as #-3\left(x-\blue{1} \right)^2+\green5 #. This means that the vertex is equal to #\rv{\blue{1},\green{5}}#.

Determine vertex with formula

The #x#-coordinate of the vertex of a quadratic #y=\blue a x^2+\green b x + \purple c# is equal to \[x_{\text{vertex}}=-\frac{\green b}{2 \blue a}\]

The #y#-coordinate of the vertex can be determined by substituting #x_{\text{vertex}}=-\frac{\green b}{2 \blue a}# in the formula.

Hence \[y_{\text{vertex}}=\blue a x^2_{\text{vertex}}+\green b x_{\text{vertex}} + \purple c\]

geogebra plaatje

Example

Take a look at the quadratic #y=\left(x+1\right)\left(x-5\right)#.

To determine the vertex we will write the formula in the form #\blue a x^2+\green b x + \purple c#. This is done in the following way:

\[\begin{array}{rcl}y&=&\left(x+1\right)\left(x-5\right) \\ &&\phantom{xxxxx}\blue{\text{the given formula}} \\&=& x^2-5x+x-5 \\ &&\phantom{xxxxx}\blue{\text{brackets expanded}}\\ &=& x^2-4x-5 \\ &&\phantom{xxxxx}\blue{\text{simplified}}\end{array}\]

Hence, we can write #y=\left(x+1\right)\left(x-5\right)# as #x^2\green{-4}x\purple{-5}#. This means that the #x#-coordinate of the vertex is equal to #x_{\text{vertex}}=-\frac{\green{-4}}{2 \cdot \blue1}=2#. To determine the #y#-coordinate of the vertex, we substitute this value in the formula:

\[y_{\text{vertex}}=\left(2+1\right)\left(2-5\right)=-9\]

Hence, the vertex is #\rv{2,-9}#.

We can prove this formula for the vertix by completing the square. This is done in the following way:

\[\begin{array}{rcl}\blue ax^2+\green bx+\purple c &=& \blue a \left(x+\frac{\green{b}}{\blue a}\right)+\purple c\\ &&\phantom{xxxxx}\blue{a \text{ expanded the brackets}}\\ &=& \blue a \left(\left(x+\frac{\green{b}}{2\blue a}\right)^2-\left(\frac{\green{b}}{2\blue a}\right)^2\right)+\purple c \\ &&\phantom{xxxxx}\blue{\text{completed the square}}\\ &=& \blue a\left(x+\frac{\green{b}}{2 \blue a}\right)^2-\blue a \cdot \left(\frac{\green b}{2\blue a}\right)^2+\purple c \\ &&\phantom{xxxxx}\blue{\text{expanded the brackets}} \\ &=& \blue a\left(x+\frac{\green{b}}{2 \blue a}\right)^2- \frac{\green b^2}{4\blue a}+\purple c \\ &&\phantom{xxxxx}\blue{\text{simplified}}\end{array}\]

Hence, the vertex is #\rv{-\frac{\green{b}}{2 \blue a}, -\frac{\green b^2}{4\blue a}+\purple c}#.

Note that the #y#-coordinate of the vertex #-\frac{\green b^2}{4\blue a}+\purple c# indeed corresponds with substituting #x_{\text{vertex}}# in #\blue ax^2+\green bx+\purple c#, want \[\begin{array}{rcl}y_{\text{vertex}}&=& \blue a \cdot \left(-\frac{\green b}{2 \blue a}\right)^2+\green b \cdot-\frac{\green b}{2 \blue a} + \purple c \\ && \phantom{xxxxx}\blue{x_{\text{vertex}}=-\frac{ b}{2 a} \text{substituted in the form‍ula }ax^2+ b x + c}\\ &=&\blue a \cdot \frac{\green b^2}{4 \blue a^2}+\green b \cdot-\frac{\green b}{2 \blue a} + \purple c \\ && \phantom{xxxxx}\blue{\text{completed the square}} \\ &=&\frac{\green b^2}{4 \blue a}-\frac{\green b^2}{2 \blue a} + \purple c \\ && \phantom{xxxxx}\blue{\text{simplified}} \\ &=&\frac{\green b^2}{4 \blue a}-\frac{2\green b^2}{4 \blue a} + \purple c \\ && \phantom{xxxxx}\blue{\text{fractions made similar}} \\&=& -\frac{\green b^2}{4\blue a}+\purple c\\ && \phantom{xxxxx}\blue{\text{fractions added}} \end{array}\]

Calculate the vertex of the formula
\[y=6-\left(x+1\right)^2\]
Give your answer in the form of #\rv{a,b}# with exact values for #a# and #b#.

#\rv{-1,6}#

The vertex of the quadratic #y=a \left(x-p\right)^2+q# is equal to #\rv{p,q}#.
Hence, the coordinates of the vertex of #y=6-\left(x+1\right)^2# is equal to #\rv{-1, 6}#.

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