Question

How do you use the important points to sketch the graph of `y= ½(x-4)(x+2)`?

Answer 1

x-intercepts `->(x,y)=(-2,0) and (4,0)`

Vertex`->(x,y)=(1,-9/2)`

`y_("intercept")->(x,y)=(0,-4)`

Explanation:

`color(blue)("Determine the general shape of the graph")`

If you multiply out the brackets you have `y=+1/2x^2+....`
As the coefficient of `x^2` is positive we have the general shape of `uu`. Thus the vertex is a minimum.

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`color(blue)("Determine the x-intercepts")`
The x-intercept is at `y=0`

This indicates that we have

`(x-4)=0=>x=+4`
`(x+2)=0=>x=-2`
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`color(blue)("Determine the vertex")`

`x_("vertex")` will be mid point between the x-intercepts so we have:

`x_("vertex")=(-2+4)/2=1`

Substitute for `x` to obtain `y_("vertex")`

`y_("vertex")=1/2(1-4)(1+2)=-9/2`

`y_("vertex")=-9/2`

Vertex`->(x,y)=(1,-9/2)`
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`color(blue)("Determine y-intercept")`

Consider the standard form `y=ax^2+bx+c`

The y-intercept is at `x=0` giving:

`y_("intercept")=c`

So from `y=1/2(x-4)(x+2)` we have:

`c=1/2xx(-4)xx2=1/2xx(-8) = -4`

`y_("intercept")->(x,y)=(0,-4)`