can someone check my work please? :)

G

Guest

Guest
Find: a) the y-intercepts and any x-intercepts, b) vertex, c) axis of symmetry.

1). y=(x+2)^2-9

a) x = 1, -5 (x+2= +/- 3) thats how i got that.
y = -5

b) vertex: -2, -9

c) axis of sym: -2


I'm pretty sure this one is right but the next one I'm not so sure.
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State the domain and range. Give the axis of symmetry. Give the vertex.

y= - x^2 - 5

Domain: (-oo, +oo)
Range: [-5, -oo)
axis of sym: 0
vertex: 0, -5

Thanks!
 
#1 You are correct. Good work.

#2 Domain: correct.
Range: Close but [-5,-oo) is the correct notation. The [-5 means that -5 is part of the range(when = = 0). (-5 means that -5 is not part of the range. -5.00...1 is as close as you can get.
Axis of sym and vertex: The axis goes thru the vertex for any parabola. You found the vertex on #1. You failed here. Did you use -b/2a to find it? The quadratic is -x²+0x-5.
 
State the domain and range. Give the axis of symmetry. Give the vertex.

y= - x^2 - 5

Domain: (-oo, +oo)
Range: [-5, -oo)
axis of sym: 0
vertex: 0, -5


i edited it with these answers, maybe i did it while you were writing, this is correct?
 
Ha, you were too fast for me :evil:
Yup, that is correct.
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Gene
 
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