How do you write #f(x) = x^2+ 8x- 5# in vertex form?

1 Answer
Jim G.
Oct 25, 2017

#f(x)=(x+4)^2-21#

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 multiplier"#

#"to obtain this form use the method of "color(blue)"completing the square"#

#• " ensure coefficient of "x^2" term is 1"#

#• " add/subtract "(1/2"coefficient of x-term")^2" to "x^2+8x#

#f(x)=x^2+8x-5larr" coefficient of "x^2" is 1"#

#color(white)(f(x))=x^2+2(4)xcolor(red)(+16)color(red)(-16)-5#

#color(white)(f(x))=(x+4)^2-21larrcolor(red)" in vertex form"#

Related questions
See all questions in Vertex Form of a Quadratic Equation
Impact of this question
8224 views around the world
You can reuse this answer
Creative Commons License