What is the vertex form of #y= x^2 – 5x – 6 #?

1 Answer
Apr 29, 2017

#y=(x-5/2)^2-49/4#

Explanation:

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

#color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))#
where (h ,k) are the coordinates of the vertex and a is a constant.

#"using the method of "color(blue)"completing the square"#

add #(1/2" coefficient of x-term")^2" to " x^2-5x#

Since we are adding a value which is not there we must also subtract this value.

#"add / subtract " (-5/2)^2=25/4#

#y=(x^2-5xcolor(red)(+25/4))color(red)(-25/4)-6#

#color(white)(y)=(x-5/2)^2-49/4larrcolor(red)" in vertex form"#