What is the vertex form of #y= 6x^2-9x+3 #?

2 Answers
Dec 14, 2017

#y = 6(x-3/4)^2 - 3/8 #

Explanation:

To complete the square of the equation, first take out the 6:

#y = 6(x^2 - 3/2x + 1/2)#

Then do the bit in the brackets:

#y = 6[(x-3/4)^2 - 9/16 + 1/2] #

#y = 6[(x-3/4)^2 - 1/16] #

#y = 6(x-3/4)^2 - 3/8 #, as required.

Dec 14, 2017

#y=6(x-3/4)^2-3/8#

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 obtain this form use the method of"#
#color(blue)"completing the square"#

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

#rArry=6(x^2-3/2x)+3#

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

#x^2-3/2x#

#rArry=6(x^2+2(-3/4)xcolor(red)(+9/16)color(red)(-9/16))+3#

#rArry=6(x-3/4)^2-27/8+3#

#rArry=6(x-3/4)^2-3/8larrcolor(red)"in vertex form"#