Question

What is the vertex form of `4y=5x^2 -7x +3`?

Answer 1

`y=color(green)(5/4)(x-color(red)(7/10))^2+color(blue)(11/80)`

Explanation:

Remember that the vertex form (our target) is in general
`color(white)("XXX")y=color(green)m(x-color(red)a)^2+color(blue)b` with vertex at `(color(red)a,color(blue)b)`

Given
`color(white)("XXX")4y=5x^2-7x+3`

We will need to divide everything by `4` to isolate `y` on the right side
`color(white)("XXX")y=5/4x^2-7/4x+3/4`

We can now extract the `color(green)m` factor from the first two terms:
`color(white)("XXX")y=color(green)(5/4)(x^2-7/5x)+3/4`

We want to write `(x^2-7/5x)` as a squared binomial by inserting some constant (which will need to be subtracted somewhere else).

Remember that the squared binomial
`color(white)("XXX")(x+p)^2=(x^2+(2p)x+p^2)`
since the coefficient of the `x` term of `(x^2-7/5x)` is `(-7/5)`
our value for `2p=-7/5 rarr p=-7/10 rarr p^2=49/100`
So we need to insert a term of `color(magenta)((-7/10)^2)=color(magenta)(49/100)` into the factor `(x^2-7/5x)` making it `(x^2-7/5+color(magenta)((-7/10)^2))`

...but remember that this factor is multiplied by `color(green)(5/4)`
so to balance thing out we will need to subtract `color(green)(5/4) xx color(magenta)(49/100)=color(brown)(49/80)`

Our equation now looks like
`color(white)("XXX")y=color(green)(5/4)(x^2-7/5+color(magenta)((-7/10)^2))+3/4-color(brown)(49/80)`

Writing this with a squared binomial and simplifying the constant terms:
`color(white)("XXX")y=color(green)(5/4)(x-color(red)(7/10))^2+color(blue)(11/80)`
which is our required vertex form with vertex at `(color(red)(7/10),color(blue)(11/80))`

For verification purposes here is a graph of the original equation:

Answer 2

`y=5/4(x-7/10)^2+11/80`

Explanation:

`"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"`

`"to express "5x^2-7x+3" in this form"`

`"use the method of "color(blue)"completing the square"`

`• "the coefficient of the "x^2" term must be 1"`

`rArr5(x^2-7/5x+3/5)`

`• " add/subtract "(1/2"coefficient of x-term")^2" to"`
`x^2-7/5x`

`5(x^2+2(-7/10)xcolor(red)(+49/100)color(red)(-49/100)+3/5)`

`=5(x-7/10)^2+5(-49/100+3/5)`

`=5(x-7/10)^2+11/20`

`rArr4y=5(x-7/10)^2+11/20`

`rArry=1/4[5(x-7/10)^2+11/20]`

`color(white)(rArry)=5/4(x-7/10)^2+11/80`