How do you graph and label the vertex and axis of symmetry of #y=-1/2(x+2)^2-1#?

1 Answer
Nov 12, 2017

#"see explanation"#

Explanation:

#"the equation of a parabola in "color(blue)"vertex form "# is.

#color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))#

#"where "(h,k)" are the coordinates of the vertex and a "#
#"is a multiplier"#

#• "if "a>0" then graph opens up"#

#• " if "a<0" then graph opens down"#

#y=-1/2(x+2)^2-1" is in "color(blue)"vertex form"#

#rArrcolor(magenta)"vertex "=(-2,-1)#

#a<0" hence graph is vertical and opens down"#

#"the axis of symmetry is therefore vertical and"#
#"passes through the vertex with equation "x=-2#

#color(blue)"Intercepts"#

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

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

#x=0toy=-1/2(2)^2-1=-3larrcolor(red)"y-intercept"#

#y=0to-1/2(x+2)^2-1=0#

#rArr(x+2)^2=-2#

#"this has no solutions hence graph has no x-intercepts"#

#color(blue)"Additional points"#

#x=2toy=-1/2(4)^2-1=-9rArr(2,-9)#

#x=-4toy=-1/2(-2)^2-1=-3rArr(-4,-3)#

#"gathering the above onto a graph gives"#
graph{(y+1/2x^2+2x+3)(y-1000x-2000)((x+2)^2+(y+1)^2-0.05)=0 [-10, 10, -5, 5]}