How do you find the vertex and the intercepts for y=3x^2+5x+9y=3x2+5x+9?

1 Answer
Jan 6, 2018

"see explanation"see explanation

Explanation:

"the equation of a parabola in "color(blue)"vertex form"the equation of a parabola in 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 y in this form use "color(blue)"completing the square"

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

rArry=3(x^2+5/3x+3)

• " add/subtract "(1/2"coefficient of x-term")^2" to"
x^2+5/3x

y=3(x^2+2(5/6)xcolor(red)(+25/36)color(red)(-25/36)+3)

color(white)(y)=3(x+5/6)^2-25/12+9

color(white)(y)=3(x+5/6)^2+83/12larrcolor(red)"in vertex form"

rArrcolor(magenta)"vertex "=(-5/6,83/12)

color(blue)"Intercepts"

• " let x = 0, in the equation for y-intercept"

• " let y = 0, in the equation for x-intercepts"

x=0toy=3(5/6)^2+83/12=9larrcolor(red)"y-intercept"

y=0to3(x+5/6)^2+83/12=0

rArr3(x+5/6)^2=-83/12

rArr(x+5/6)^2=-83/36

"this has no real solutions hence no x-intercepts"
graph{3x^2+5x+9 [-20, 20, -10, 10]}