How do you find the axis of symmetry, vertex and x intercepts for #y=3x^2+12x+5#?

1 Answer
Dec 17, 2017

#x=-2,(-2,-7),x=-2+-sqrt(7/3)#

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"#

#rArry=3(x^2+4x)+5#

#• " add/subtract "(1/2"coefficient of x-term")^2" to"#
#x^2+4x#

#rArry=3(x^2+2(2)x color(red)(+4)color(red)(-4))+5#

#color(white)(rArry)=3(x+2)^2-12+5#

#color(white)(rArry)=3(x+2)^2-7larrcolor(red)"in vertex form"#

#rArrcolor(magenta)"vertex "=(-2,-7)#

#"the axis of symmetry passes through the vertex is vertical"#
#"with equation"#

#x=-2larrcolor(blue)"axis of symmetry"#

#"to obtain "color(blue)"x-intercepts" " set y = 0"#

#rArr3(x+2)^2-7=0#

#rArr(x+2)^2=7/3#

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

#sqrt((x+2))^2=+-sqrt(7/3)larrcolor(blue)"note plus or minus"#

#rArrx+2=+-sqrt(7/3)#

#"subtract 2 from both sides"#

#rArrx=-2+-sqrt(7/3)larrcolor(blue)"exact values"#

#rArrx~~-3.53" or "x~~-0.47larrcolor(red)"x-intercepts"#
graph{(y-3x^2-12x-5)(y-1000x-2000)=0 [-20, 20, -10, 10]}