How do you find the vertex and the intercepts for #y= x^2-8x+5#?
1 Answer
Jul 21, 2018
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(-4)x+16-16+5#
#color(white)(y)=(x-4)^2-11larrcolor(blue)"in vertex form"#
#color(magenta)"vertex "=(4,-11)#
#"to obtain the y-intercept let x = 0"#
#y=0-0+5=5larrcolor(red)"y-intercept"#
#"to obtain the x-intercepts let y = 0"#
#(x-4)^2-11=0#
#(x-4)^2=11#
#color(blue)"take the square root of both sides"#
#x-4=+-sqrt11larrcolor(blue)"note plus or minus"#
#"add 4 to both sides"#
#x=4+-sqrt11larrcolor(red)"exact solutions"#
#x~~0.68" or "x~~7.32" to 2 dec. places"#