How do you find the vertex and the intercepts for #f(x)=-x^2+2x+4#?

1 Answer
Jul 30, 2018

#"vertex "=(1,5),x=1+-sqrt5#

Explanation:

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

#•color(white)(x)y=a(x-h)^2+k#

#"where "(h,k)" are the coordinates of the vertex and a"#
#"is a multiplier"#

#"to obtain this form "color(blue)"complete the square"#

#y=1(x^2+2(-1)x+1-1-4)#

#color(white)(y)=-(x-1)^2+5#

#color(magenta)"vertex "=(1,5)#

#"let x = 0 for y-intercept"#

#y=-1+5=4larrcolor(red)"y-intercept"#

#"let y = 0 for x-intercepts"#

#-(x-1)^2+5=0#

#(x-1)^2=5#

#x-1=+-sqrt5#

#x=1+-sqrt5larrcolor(red)"exact values"#