#"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]}
