"given a parabola in standard form ";y=ax^2+bx+c
"then the x-coordinate of the vertex can be found using"
•color(white)(x)x_(color(red)"vertex")=-b/(2a)
f(x)=x^2+4x-5" is in standard form"
"with "a=1,b=4,c=-5
rArrx_(color(red)"vertex")=-4/2=-2
"substitute this value into "f(x)" for y coordinate"
y_(color(red)"vertex")=(-2)^2+4(-2)-5=-9
rArrcolor(magenta)"vertex "=(-2,-9)
"to obtain the intercepts"
• " let x = 0, in equation for y-intercept"
• " let y = 0, in equation for x-intercepts"
x=0toy=-5larrcolor(red)"y-intercept"
y=0tox^2+4x-5=0
"the factors of - 5 which sum to + 4 are + 5 and - 1"
rArr(x+5)(x-1)=0
"equate each factor to zero and solve for x"
x+5=0rArrx=-5
x-1=0rArrx=1
x=-1" and "x=-5larrcolor(red)"x-intercepts"
graph{x^2+4x-5 [-20, 20, -10, 10]}
