How do you find the vertex and intercepts for #9x^2 – 12x + 4 = 0 #?

1 Answer
Apr 25, 2018

#(2/3,0)" and "x=2/3" (repeated)"#

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 "color(blue)"completing the square"#

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

#"factor out 9"#

#9(x^2-4/3x+4/9)=0#

#• " add/subtract "(1/2"coefficient of the x-term")^2" to"#
#x^2-4/3x#

#9(x^2+2(-2/3)xcolor(red)(+9/4)color(red)(-9/4)+9/4)=0#

#rArr9(x-2/3)^2+0=0#

#"the left side is now in "color(blue)"vertex form"#

#"with "h=2/3" and k=0#

#rArrcolor(magenta)"vertex "=(2/3,0)#

#"for intercepts solve the equation "#

#rArr9(x-2/3)^2=0#

#rArrx=2/3"( repeated)"#

#"this indicates a minimum at "(2/3,0)#
graph{9x^2-12x+4 [-10, 10, -5, 5]}