Question

find the domain and range of

Answer 1

2
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x^2-2x-3

Answer 2

Can you factor the denominator?

Answer 3

can you show me please?

Answer 4

x^2 - 2x - 3 = (x - 3)(x + 1)

Right?

Answer 5

ok

Answer 6

That stuff is in the denominator, and you probably know you can't divide by zero, right?

So if either of those factors in parenthesis is equal zero, the denominator is zero, and that buzzer error noise should go off in your head.

Answer 7

You need to restrict the domain (the input values) so that you don't include values which cause "divide by zero" errors.

Answer 8

when x = 3 , the first parenthesis solves to zero, and when x = -1, the second parenthesis solves to zero.

So the domain is all real x values, EXCEPT for x=3 and x =-1.

Answer 9

The range (output values) is just the expression, but you need to say the range is "undefined" for x=3 and x=-1

Answer 10

so i write the range is undefined for x=3 and x=-1
and how would you find the domain using a graph

Answer 11

I don't know... do you have a graph of the expression?

Answer 12

how would you graph it on paper?

Answer 13

It looks messy... I would make a table with x in the left column, put in different x values and solve, putting the answers in the right column, then plot points.

Answer 14

The areas of interest though are x values below and above -1 and below and above 3

These are the two points that aren't "allowed", so look at the graph in areas around and in between those points

Answer 15

plot the x value and the resulting value from solving 2 / [ (x+1)(x-3) ] for each value of x

Answer 16

I don't think I would really recommend doing that though unless you are just curious what the curve looks like. You don't need it to determine domain and range... the point of the question is to make you realize that some x values will cause "divide by zero" errors, therefore those values are not part of the domain

Answer 17

i have to graph that expression on paper and take a picture of it

Answer 18

wow. that's no fun... Just like you would make a graph of a line by putting in x values, solving for the y, and then plotting the (x,y) points, do the same here... think of an x value, evaluate that expression using that x, and plot the point. Repeat for lots of points. At x = -1 and x = 3, leave an empty circle in the curve to show that those points aren't included

Answer 19

what are the x and y Intercept(s)?

Answer 20

Something like this, except I don't know the real shape... I just made this up