How do you find the vertex and intercepts for #f(x)=-3x^2+5x+9#?

1 Answer
Nov 29, 2017

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

#• " ensure the coefficient of the "x^2" term is 1"#

#rArrf(x)=-3(x^2-5/3x-3)#

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

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

#color(white)(f(x))=-3(x-5/6)^2+133/12larrcolor(blue)"in vertex form"#

#rArrcolor(magenta)"vertex "=(5/6,133/12)#

#color(blue)"Intercepts"#

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

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

#x=0toy=9larrcolor(red)"y-intercept"#

#y=0to-3(x-5/6)^2+133/12=0#

#rArr-3(x-5/6)^2=-133/12#

#rArr(x-5/6)^2=133/36#

#color(blue)"take the square root of both sides"#

#rArrx-5/6=+-sqrt(133/36)larrcolor(blue)"note plus or minus"#

#rArrx=5/6+-sqrt133/6larrcolor(red)"exact solutions"#

#rArrx~~-1.09" or "x~~2.76larrcolor(red)"x-intercepts"#