Question

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

Answer 1

`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)`

Answer 2

`(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]}

Answer 3

`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"))`