# 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"#