Question

What is the vertex form of `y= (5x-9)(3x+4)+x^2-4x`?

Answer 1

See below.

Explanation:

First multiply out the brackets and collect like terms:

`15x^2 - 27x + 20x - 36 + x^2 - 4x => 16x^2 - 11x - 63`

Bracket terms containing the variable:

`( 16x^2 - 11x ) - 63`

Factor out the coefficient of `x^2`:

`16( x^2 - 11/16x ) - 63`

Add the square of half the coefficient of `x` inside the bracket, and subtract the square of half the coefficient of `x` outside the bracket.

`16( x^2 - 11/16x +( 11/32)^2 ) - 63 - (11/32)^2`

Rearrange `( x^2 - 11/16x +( 11/32)^2 )` into the square of a binomial.

`16( x - 11/32 )^2 - 63 - ( 11/32)^2`

Collect like terms:

`16( x - 11/32 )^2 - 63 - ( 11/32)^2`

`16( x - 11/32 )^2 - 64633/1024`

This is now in vertex form: `a( x - h )^2 + k`
Where `h` is the axis of symmetry and `k` is the maximum or minimum value of the function.

So from example:

`h = 11/32` and `k = -64633/1024`

Answer 2

`y=16(x-11/32)^2-2425/64`

Explanation:

`"the first step is to rearrange the parabola in standard form"`

`"that is "y=ax^2+bx+cto(a!=0)`

`"expand factors using FOIL and collect like terms"`

`y=15x^2-7x-36+x^2-4x`

`color(white)(y)=16x^2-11x-36larrcolor(red)" in standard form"`

`"the x-coordinate of vertex in standard form is"`

`x_(color(red)"vertex")=-b/(2a)`

`y=16x^2-11x-36`

`"with "a=16,b=-11,c=-36`

`rArrx_(color(red)"vertex")=-(-11)/(32)=11/32`

`"substitute this value into the equation for y"`

`y_(color(red)"vertex")=16(11/2)^2-11(11/32)-36=-2425/64`

`rArrcolor(magenta)"vertex "=(11/32,-2425/64)`

`"the equation of a parabola in "color(blue)"vertex form"` is.

`color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))`
where )h , k ) are the coordinates of the vertex and a is a multiplier.

`"here "(h,k)=(11/32,-2425/64)" and "a=16`

`rArry=16(x-11/32)^2-2425/64larrcolor(red)" in vertex form"`