Question

How do you find the intercept and vertex of `y = (x − 3)(4x + 2)`?

Answer 1

Solution Part 1 of 2

`x_("intercepts")-> x=3 and x=-1/2`

`y_("intercept")=-6`

Vertex`->(x,y)=(5/4,-49/4)`

Explanation:

`color(blue)("Determine the x-intercepts")`

x-intercept is at `y=0`

Set `y=0=(x-3)(4x+2)`

Consider the case `(x-3)=0 => x=3`
Consider the case `(4x+2)=0->2(2x+1)=0=>x=-1/2`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the vertex")`

If you multiply out the brackets you have `+4x^2 ....`. This means that the graph is of form `uu` thus the vertex is a minimum.

`x_("vertex")` will be half way between the x-intecpts

`x_("vertex")=(3+(-1/2))/2 = 2.5/2=1.25 -> 5/4`

Thus `y_("vertex") = (5/4-3)(5+2) = -7/4xx7=-49/4`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(blue)("Determine the y-intercept")`

Multiplying out the constants in the brackets we have

`(-3)xx(+2) = -6`

Thus `y_("intercept")=-6`

Answer 2

Answer part 2 of 2

Building 'completion of the square'

Explanation:

Notice that the section for this question is 'Vertex Form ....'

`y=(x-3)(4x+2)=4x^2-10x-6`

Set `y=4(x^2-10/4x)-6`

Halve the coefficient of `x`

`y=4(x^2-10/8x)-6`

Remove the `x` from `-10/8x` and move the squaring to outside the brackets

`y=4(x-10/8)^2-6` This will not produce the original equation `color(white)("ddddddddddddddddd")`as we have not includes a correction
`color(white)("dddddddddddddddddd")`process.

`y=color(red)(4)(xcolor(green)(-10/8))^2+k-6` Now it will!

Note that `color(red)(4)xxcolor(green)((-10/8)^2)+k=0`

`=> k=-25/4`

So `k-6->-25/4-24/4=-49/4`

So the final format is:

`y=4(x-5/4)^2-49/4`