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

2 Answers
Jan 12, 2018

#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:
enter image source here

Jan 12, 2018

#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#