How do you find the vertex and the intercepts for #f(x)=-x^2+2x-6#?

1 Answer
Sep 4, 2017

#"see explanation"#

Explanation:

#"for a parabola in standard form "f(x)=ax^2+bx+c#

#"the x-coordinate of the vertex is "#

#x_(color(red)"vertex")=-b/(2a)#

#f(x)=-x^2+2x-6" is in standard form"#

#"with "a=-1,b=2,c=-6#

#rArrx_(color(red)"vertex")=-2/(-2)=1#

#"substitute this value into the equation for y"#

#y_(color(red)"vertex")=-1^2+2-6=-5#

#rArrcolor(magenta)"vertex "=(1,-5)#

#color(blue)"Intercepts"#

#• " let x = 0, in the equation for y-intercept"#

#• " let y = 0, in the equation for x-intercepts"#

#x=0toy=-6larrcolor(red)" y-intercept"#

#y=0to-x^2+2x-6=0#

#"check the "color(blue)"discriminant"#

#Delta=b^2-4ac=4-24=-20#

#Delta<0rArr" equation has no real solutions"#

#"hence there are no x-intercepts"#
graph{-x^2+2x-6 [-16.02, 16.02, -8, 8.02]}