How do you find the vertex and the intercepts for #f(x) = -x^2 + 2x + 5#?

2 Answers
Jun 6, 2017

it vertex is maximum at #(1,6)#

x-intercepts # = 1 +- sqrt6#

y-intercept # = 5#

Explanation:

#f(x) = -x^2 + 2x + 5#

#f(x) = -(x^2 -2x) + 5#

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

#f(x) = -(x -1)^2 +6#

since coefficient of #(x-1)# is -ve value, it vertex is maximum at #(1,6)#

to find x-intercept, plug in #f(x) =0# in the equation, therefore
#0 = -(x -1)^2 +6#
#(x -1)^2 = 6#
#x -1 = +- sqrt6#
#x = 1 +- sqrt6#

to find y-intercept, plug in #x =0# in the equation, therefore
#f(0) = -(0-1)^2 +6#
#f(0) = -( -1)^2 +6#
#f(0) = -1 +6 = 5#

Jun 6, 2017

#"see explanation"#

Explanation:

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

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

#y=-x^2+2x+5" is in this form"#

#"with " a=-1,b=2" and " c=5#

#rArrx_(color(red)"vertex")=-2/(-2)=1#

#"substitute this value into function for y-coordinate"#

#rArry_(color(red)"vertex")=-(1)^2+(2xx1)+5=6#

#rArrcolor(magenta)"vertex "=(1,6)#

#color(blue)" for intercepts"#

#• " let x = 0, in function for y-intercept"#

#• " let y = 0, in function for x-intercepts"#

#x=0toy=0+0+5=5larrcolor(red)" y-intercept"#

#y=0to-x^2+2x+5=0#

#"solve using the "color(blue)"quadratic formula"#

#x=(-2+-sqrt(4+20))/(-2)=(-2+-sqrt24)/-2=(-2+-2sqrt6)/-2#

#rArrx=1+-sqrt6#

#rArrx~~3.45,x~~-1.45larrcolor(red)" x-intercepts"#
graph{-x^2+2x+5 [-12.65, 12.66, -6.34, 6.3]}