How do you find the vertex of #y= x^2 + 8x -9#?

2 Answers
May 20, 2018

#"vertex "=(-4,-25)#

Explanation:

#"given a parabola in "color(blue)"standard form"#

#•color(white)(x)y=ax^2+bx+c color(white)(x);a!=0#

#"then the x-coordinate of the vertex is"#

#•color(white)(x)x_(color(red)"vertex")=-b/(2a)#

#y=x^2+8x-9" is in standard form"#

#"with "a=1,b=8" and "c=-9#

#rArrx_("vertex")=-8/2=-4#

#"substitute this value into the equation for y"#

#y_("vertex")=(-4)^2+8(-4)-9=-25#

#rArrcolor(magenta)"vertex "=(-4,-25)#

May 20, 2018

The vertex is #(-4,-25)#

Explanation:

You can change the standard form which is given into vertex form which is #y = a(x+p)^2+q# by the process of completing the square. The vertex is #(-p,q)#

#y=x^2 +8x -9#

To complete the square, add and subtract #color(blue)((b/2)^2 =16)#
#y = x^2 +8x " " color(blue)( +16 -16) -9#

#y = (x^2 +8x color(blue)(+16)) +(color(blue)(-16) -9)#

#y = (x+4)^2-25#

The vertex is #(-4,-25)#