Home > Precalculus > Asymptotes

How do you find the zeros and holes, if any, of #R( x) = (3x)/(x^2 - 2x)#?

Answer 1

Elijah Allen

#"hole at " x=0, " there are no zeros"#

#"factorise and simplify " R(x)#

#R(x)=(3cancel(x))/(cancel(x)(x-2))=3/(x-2)#

Since we have removed a factor of x, this indicates there is a hole at x=0

Zeros are the values of x that make R(x) equal zero.

This occurs when the numerator is zero.

#"Since the numerator is 3 there are no zeros"# graph{(3x)/(x^2-2x) [-10, 10, -5, 5]}

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Caleb Alexander

To find the zeros and any holes of the rational function ( R(x) = \frac{3x}{x^2 - 2x} ), first, factor the denominator. Then, set the numerator equal to zero to find any zeros, and check for common factors to identify any holes in the function.

Factoring the denominator gives ( x(x - 2) ). Setting the numerator equal to zero gives ( 3x = 0 ), so ( x = 0 ) is a zero of the function. To find if there are any holes, check for common factors between the numerator and denominator. Since there are no common factors, there are no holes in this function. Therefore, the only zero of the function is ( x = 0 ), and there are no holes.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.