Question

PLEASE HELP ME!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
I WILL AWARD MEDAL!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Simplify the following equation and find the restriction
y^2 -y -6
--------
y^2 -4

Answer 1

@Mertsj @zscdragon @Mousam

Answer 2

same thing
factorise the top and the bottome and cancel out the similar terms

Answer 3

Ok it is y-3
----
y-2
But the restriction is the problem.

Answer 4

You mean the restricted values? They'll be values that make the denominator equal to zero (you can't divide by zero)

You need to solve for y \[\Large y^2 - 4 = 0\]

Answer 5

I know that it is just I got it wrong last tim.

Answer 6

Can you factor y^2-4? you should get \[\large (y+2)(y-2) = 0\]Now solve that.

Answer 7

Oh so the restriction is pluse or minus 2

Answer 8

and z =1 too

Answer 9

Thank you so much so I am right,

Answer 10

Er,, there's no z?

Answer 11

yes no Z

Answer 12

where z = y^2 -y -6
--------
y^2 -4

Answer 13

Restricted values are concerned with the domain, not range.

Answer 14

trust me
or if you font trust me.. try a graphic calculator

Answer 15

And there's no z. The equation at the start was

y^2 -y -6
--------
y^2 -4

Answer 16

ok...

Answer 17

z=1 is not a restricted value even if
z = y^2 -y -6
--------
y^2 -4

Restricted values are values in the domain, not range.

Answer 18

I still have a few more questions any one willing to help me.

Answer 19

@Mousam you're talking about a horizontal asymptote. Not a restricted value.

Answer 20

i thought
\[z=\frac{ y-3 }{ y-2 }\]\[\lim_{y \rightarrow \infty} z=\frac{ 1-\frac{ 3 }{ y } }{ 1-\frac{ 2 }{ y } } = \]

Answer 21

oh so he is just asking for the vertical ones?

Answer 22

Again, restricted values are values not in the domain. y=2 and y=-2 are not in the domain.

y=2 is also a vertical asymptote. y=-2 is a point of discontinuity.

z=1 is a horizontal asymptote.