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]}