How do you find the intercept and vertex of #y = x^2 - 4x - 2#?

1 Answer
Jul 24, 2018

#"vertex "=(2,-6)," intercepts "=2+-sqrt6#

Explanation:

#"The equation of a parabola in "color(blue)"vertex form"# is.

#•color(white)(x)y=a(x-h)^2+k#

#"where "(h,k)" are the coordinates of the vertex and a"#
#"is a multiplier"#

#"to obtain this form "color(blue)"complete the square"#

#y=x^2+2(-2)x+4-4-2#

#y=(x-2)^2-6larrcolor(red)"in vertex form"#

#color(magenta)"vertex "=(2,-6)#

#"to obtain the x-intercepts let y = 0"#

#(x-2)^2-6=0#

#(x-2)^2=6#

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

#x-2=+-sqrt6larrcolor(blue)"note plus or minus"#

#"add 2 to both sides"#

#x=2+-sqrt6larrcolor(red)"exact values"#