How do you find the vertex and intercepts for #y+12x-2x^2=15#?

1 Answer
Dec 12, 2017

#"see explanation"#

Explanation:

#"express in standard form"#

#rArry=2x^2-12x+15#

#"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"#

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

#rArr2(x^2-6x)+15#

#• " add/subtract "(1/2"coefficient of x-term")^2"to"#
#x^2-6x#

#2(x^2+2(-3)xcolor(red)(+9)color(red)(-9))+15#

#=2(x-3)^2-18+15#

#rArry=2(x-3)^2-3larrcolor(red)"in vertex form"#

#rArrcolor(magenta)"vertex "=(3,-3)#

#color(blue)"Intercepts"#

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

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

#x=0toy=2(-3)^2-3=15larrcolor(red)"y-intercept"#

#y=0to2(x-3)^2-3=0#

#rArr(x-3)^2=3/2#

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

#rArrx-3=+-sqrt(3/2)larrcolor(blue)"note plus or minus"#

#rArrx=3+-sqrt(3/2)larrcolor(red)"x-intercepts"#

#rArrx~~1.78,x~~4.22" to 2 dec. places"#