How to find the domain of the function y=5/x-3?

Question

and if anyone could explain how to do this on a graphing calculator that would be great

Answer 1

when you go to y-equals and graph the equation, then the graph should look like an upside down and backward L to the left of the 3 and a normal L to the right of the 3. In between the two should be an almost vertical line, this line is called the asymptote. That means that the domain of the function is all reals, except for 3, since there is an asymptote there...

Answer 2

x can be any real number except 3 because x=3 would cause division by zero which is not allowed and is undefined.
So the domain is (-infinity,3) U (3, =infinity). This notation simply means x can be any real value except 3.

If you graph y= 5/(x-3) on a calculator, you will get a verical asymptote at x = 3. This means that x can get as close as you like to 3 but can never reach it. The calculator should show you a vertical line at x=3.

Answer 3

For the area: If the function is composed of fractions, then set the denominator equivalent to 0 and remedy for the variable. occasion: #2 h(x) = 5/(x-3) x-3 = 0 x = 3 area = all genuine numbers different than 3. If the function is composed of a sq. root, then set despite is below the sq. root greater advantageous than or equivalent to 0 and remedy for the variable. this may be your area. occasion: #8 f(x) = sqrt(x^4 - 16x^2) x^4 - 16x^2 >= 0 x^2(x^2 - sixteen) >= 0 x^2(x-4)(x+4) >= 0 Your severe factors take place while x^2 = 0, x-4 = 0 or x+4 = 0 x = 0, x = 4 or x = -4 x^4 - 16x^2 is larger than or equivalent to 0 while x <= -4, x = 0, x>= 4 area = (-infinity,-4] U {0} U [4,infinity)

Answer 4

all reals except 3

Answer 5

answer in the 3rd quadrant. lower left .