How do you find the vertex of #y= x^2 + 8x -9#?
2 Answers
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)#
The vertex is
Explanation:
You can change the standard form which is given into vertex form which is
To complete the square, add and subtract
The vertex is