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]}