Could you find the x-intercepts of these two rational funtions?

Question

Please find the x- intercepts of the following two functions:

a) (x^2-4)/((x^2-9)

and

b) (x^3-6x+5)/(x^2-3)

Thank you

Answer 1

a) Equal the funtion to zero in order to get intercepts
(x+2)(x-2)
------------ = 0
(x+3)(x-3)

x= 2
x=-2

b) Factorizing and solving

(x-1) (x² +x -5)
--------------------=0
x² - 3

We should use the general formula to get the second term result

x= 1 is the first root

now for x² +x -5=0

x= -1 ± √21 / 2

x= (-1 + √21)/2 second root

x= (-1 - √21)/2 third root

Answer 2

Graphing calculator???!!! They should be banned from math class. In physics or engineering they are fine, but in math you are trying to understand what is happening, using a magic box doesn't help.

a) Factor: (x + 2)(x - 2)/((x + 3)(x - 3)

x intercepts come where y is 0, so this is at +-2. There are vertical asymptotes where the denominator is 0, at +-3. The graph is positive > 3 and < -3, on the intervals (-3,-2) and (2,3) it is negative, and (-2,2) is positive.

(x^3 - 6x + 5)/(x^2 - 3): by inspection x = 1 is a zero, so divide (x^3 - 6x + 5)/(x - 1) gives (x^3 - 6x + 5) = (x - 1)(x^2 + x - 5)

Using the quadratic equation to solve the other factor:

x = (-1 +- sqrt(1 + 20))/2

So unless I made a mistake in the division, there are 3 zeros, at the points 1, -1/2 + sqrt(21)/2, and -1/2 - sqrt(21)/2

Answer 3

ok X squared minus 4 right? (X+2) (X-2) plus and minus 2 for A

X^3 - 6x +5 = 0 just put in -1 and you get zero those are the answers

Answer 4

To find x-intercepts, set the function equal to 0
a)
y = (x^2-4)/((x^2-9)
(x + 2)(x - 2) = 0
x = -2, 2

and

b)
y = (x^3-6x+5)/(x^2-3)

..... . x^2 + x - 5
x - 1)x^3..... . . - 6x + 5 = 0
.......-x^3 + x^2
..... ..... . + x^2 - 6x
..... ....... - x^2.. + x
..... ..... ..... .... - 5x + 5
..... ..... ..... .... + 5x - 5

x = 1, (-1 ± √(1 + 20))/2
x = 1, (-1 ± √21)/2
x = 1, -2.7913, 1.7193

Answer 5

Factorise:
a. (x+2)(x-2)/[(x+3)(x-3)]. Roots: x = -2, 2

Answer 6

Get yourself a graphing calculator. They used to be mandatory for algebra/calculus when I was in school. They can solve equations and you can examine graphs. Granted you are supposed to do this by hand.