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

1 Answer
Jan 6, 2018

#"see explanation"#

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 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]}