What is the the vertex of #y = 1/2(x+1)(x-5) #?

3 Answers
Dec 13, 2015

#y= 1/2 (x-color(red)(2))^2 color(blue)(-9/2)#

vertex: #(2, -9/2)#

Explanation:

Note:

Vertex form #f(x) = a(x-h)^2+k#
#h= x_(vertex) = -b/(2a) " " " "# ; #k= y_(vertex)= f(-b/(2a))#

Given:

#y= 1/2 (x+1)(x-5)#

Multiply the expression or FOIL

#y = 1/2 (x^2 -5x+x-5)#

#y = 1/2(x^2 -4x-5)#

#y= 1/2x^2 -2x -5/2#

#a = 1/2;" " b= -2;" " " c= -5/2#

#color(red)(h= x_(vertex)) = (-(-2))/(2*1/2) =color(red) 2#
#color(blue)(k= y_(vertex)) = f(2) = 1/2(2)^2 -2(2) -5/2 #

#=> 2-4 -5/2 => -2 -5/2 => color(blue)(-9/2 #

The vertex form is

#y= 1/2 (x-color(red)(2))^2 color(blue)(-9/2)#

Dec 13, 2015

#(2,-9/2)#

Explanation:

First, find the expanded form of the quadratic.

#y=1/2(x^2-4x-5)#

#y=1/2x^2-2x-5/2#

Now, the vertex of a parabola can be found with the vertex formula:

#(-b/(2a),f(-b/(2a)))#

Where the form of a parabola is #ax^2+bc+c#.

Thus, #a=1/2# and #b=-2#.

The #x#-coordinate is #-(-2)/(2(1/2))=2#.

The #y#-coordinate is #f(2)=1/2(2+1)(2-5)=-9/2#

Thus, the vertex of the parabola is #(2,-9/2)#.

You can check the graph:
graph{1/2(x+1)(x-5) [-10, 10, -6, 5]}

Dec 14, 2015

#color(blue)("A slightly quicker approach")#

#color(green)("It is not uncommon for there to be several ways of solving a problem!")#

Explanation:

This is a quadratic thus of the hors shoe type shape.

That means that the vertex is #1/2# way between the x-intercepts.

The x-intercepts will occur when y=0

If y is 0 then the right side also = 0

The right side equals zero when #(x+1)=0 " or " (x-5)=0#

For #(x+1)=0 -> x=-1#
For#(x-5)=0 -> x =+5#

Half way is #((-1)+(+5))/2 = 4/2=2#

Having found #color(blue)(x_("vertex")=2)# we then substitute in the original equation to find #color(blue)(y_("vertex"))#