How do you find the vertex and the intercepts for #f(x) = -x^2 + 2x + 5#?
2 Answers
it vertex is maximum at
x-intercepts
y-intercept
Explanation:
since coefficient of
to find x-intercept, plug in
to find y-intercept, plug in
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]}