Question

Find the intercepts of the graph of the equation.
x^2-9/x^2-4

Answer 1

when x=0, what does y equal?

when y=0, what does x need to equal? hint, the top goes zero

Answer 2

So I do 9/4?

Answer 3

that will be the y intercept, yes

Answer 4

and recall that in order for a fraction to be zero; then the top part has to be zero ... as long as the bottom part doesnt zero out as well

Answer 5

2.25

Answer 6

for the y intercept, yes. 2.25

Answer 7

I think I am doing it wrong...

Answer 8

so (0,2.25)

Answer 9

y = x^2-9/x^2-4

9/4 = 0^2-9/0^2-4 = 2.25


y = x^2-9/x^2-4

0 = 0/x^2-4, so for this we will need to determine x^2-9 = 0

Answer 10

What do I plug in?

Answer 11

umm, for the second one? you plug in y=0 and solve for x

Answer 12

x^2 - 9 = 0 , when x=?

Answer 13

0

Answer 14

lets see:

0^2 - 9 not= 0 , so x=0 is not correct

Answer 15

-9

Answer 16

ummm, lets see

(-9)^2 - 9 = 0

81 - 9 ... not equal 0, so x=-9 is not it

Answer 17

A plot may help.

Answer 18

x^2 - 9 = 0 , lets start by adding 9 to each side

x^2 -9 +9 = 0 + 9

x^2 = 9 , lets take the square root of each side to undo that ^2

sqrt(x^2) = sqrt(9) , lets take the square root of each side to undo that ^2

x = +- sqrt(9)

Answer 19

and since: sqrt(9)^2 - 4 not equal 0, the denominator is safe .....

Answer 20

x= to 0 though

Answer 21

not sure i follow you on that ine

Answer 22

I am confused sorry

Answer 23

what do you mean the denominator is safe?

Answer 24

we are not allowed to divide by zero in math .... so as long as a solution for the top part does not affect the bottom part of the fraction, we are good to go.

Answer 25

oh ok

Answer 26

\[\frac{x^2-9}{x^2-4}\frac{\text{<-- needs to be zero}}{\text{<-- cant be 0}}\]

Answer 27

\[3^2-9=0\]\[3^2-4\ne0\]safe

\[(-3)^2-9=0\]\[(-3)^2-4\ne0\]safe

Answer 28

So how do I find the intercepts?

Answer 29

so y=3?

Answer 30

or x=3

Answer 31

....

to find the y intercept, make x=0: (0,y)

to find the x intercept (may be many of them), make y=0: \((x_1,0),(x_2,0),...\)

Answer 32

(0,3)

Answer 33

\[y=\frac{x^2-9}{x^2-4}\]
yint is x=0
\[y=\frac{0^2-9}{0^2-4}=\frac94~;~yint~(0,\frac94)\]


xint is when y=0
\[0=\frac{x^2-9}{x^2-4}\]since a fraction is 0 when the top is zero ...\[0=\frac{0}{x^2-4}~;~(x^2-9=0)~;~(x^2-4)\ne0\]

Answer 34

xints are therefore (3,0) and (-3,0)

Answer 35

So they basically have to equal

Answer 36

no, thats just a coincidence of this particular setup. It not a general conclusion

Answer 37

if im reading you correctly that is ...

Answer 38

thank you

Answer 39

good luck