How do you graph the quadratic function and identify the vertex and axis of symmetry for #y=-(x-2)^2-1#?
1 Answer
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 constant.
#y=-(x-2)^2-1" is in this form"# and by comparison
#h=2" and " k=-1#
#rArr"vertex "=(2,-1)# Since the
#color(blue)"value of a is negative"#
#"That is " color(red)(-)(x-2)^2# Then the parabola opens down vertically
#color(red)(nnn# The axis of symmetry goes through the vertex and therefore has equation
#color(magenta)"x=2"#
#color(blue)"Intercepts"#
#x=0toy=-(0-2)^2-1=-4-1=-5#
#rArry=-5larrcolor(red)"y-intercept"#
#y=0to-(x-2)^2-1=0to(x-2)^2=-1# This has no real solutions hence there are no x-intercepts.
Knowing the ' shape' of the parabola, the vertex and the y-intercept enables the graph to be sketched.
graph{-(x-2)^2-1 [-10, 10, -5, 5]}
