# 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]}