The domain of a rational function can never be the set of all real numbers. true or false. justify your answer. true or false. justify your answer. the graph of a polynomial function can have infinitely many vertical asymptotes. true or false? justify your answer.

"the domain of a rational function can never be the set of all real numbers."

this is a false statement.

a rational function is defined to be quotient of two polynomial expressions.

what makes a rat. func. undefined is that there is a value that makes the denominator equal to zero; you are not allowed to divide by 0.

so consider the rational function

f(x) = x/(x^2+1)

the denominator is never 0 as the smallest number the denominator can be is 1 for x = 0. therefore, the domain of f is all real numbers; there's no undefined value. we have found a counter-example that disproves that conjecture.

an easier counter-example to use is the function f(x) = 1. this is a quotient of two degree-0 polynomials, , and qualifies as a rational function. its domain is all real numbers.

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"the graph of a polynomial function can have infinitely many vertical asymptotes."

this is a false statement.

by definition, all polynomial functions are continuous and have their domains as all real numbers. if a graph of a function has a vertical asymptote, that implies that the x-coordinate of that asymptote is not in the domain of that function. this is a contradiction to the definition of a polynomial function.

polynomial functions cannot have vertical asymptotes, therefore it cannot have infinitely many asymptotes.

5x+40

5(x + 8)

7x - 49

7(x - 7)

18x + 9

9(2x + 1)

15x - 10

5(3x - 2)

step-by-step explanation:

for number one all you have to do is factor 5 out of 5x + 40

for number 2 factor 7 out of 7x - 49

for number three factor 9 out of 18x + 9

for number four factor 5 out of 15x - 10