How do you find the maximum or minimum of #f(x)=-1/2x^2-2x+3#?

1 Answer
Dec 15, 2017

#(-2,5)" is the maximum"#

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"#

#• " if "a>0" then f(x) is a minimum"#

#• " if "a<0" then f(x) is a maximum"#

#(h,k)" will be the max/min turning point of f(x)"#

#"to express "f(x)=-1/2x^2-2x+3" in vertex form"#

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

#• " the coefficient of the "x^2" term must be 1"#

#rArrf(x)=-1/2(x^2+4x)+3#

#• " add/subtract "(1/2"coefficient of x-term")^2" to"#
#x^2+4x#

#f(x)=-1/2(x^2+2(2)xcolor(red)(+4)color(red)(-4))+3#

#color(white)(f(x))=-1/2(x+2)^2+2+3#

#color(white)(f(x))=-1/2(x+2)^2+5larrcolor(blue)"in vertex form"#

#rArrcolor(magenta)"vertex "=(-2,5)" and "a<0#

#rArr"maximum turning point "=(-2,5)#
graph{(y+1/2x^2+2x-3)((x+2)^2+(y-5)^2-0.04)=0 [-10, 10, -5, 5]}