How do you write # y =x^2 - 5x - 6# in vertex form?

1 Answer
Jim G.
Aug 7, 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.

#"for a parabola in standard form "y=ax^2+bx+c#

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

#y=x^2-5x-6" is in standard form"#

#"with "a=1,b=-5,c=-6#

#rArrx_(color(red)"vertex")=-(-5)/2=5/2#

#"substitute this into the equation for y-coordinate"#

#rArry_(color(red)"vertex")=(5/2)^2-5(5/2)-6=-49/4#

#rArrcolor(magenta)"vertex "=(5/2,-49/4)#

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

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