We set the denominator to zero to find the singularities: points where the graph is undefined.
The function is not defined at any values at which the denominator is zero.
Rational expressions are fractions and are therefore undefined if the denominator is zero; the domain of a rational function is all real numbers except those that make the denominator of the related rational expression equal to 0. If a denominator contains variables, set it equal to zero and solve.
True
It cannot be zero.
It is a rational fraction.
The function is not defined at any values at which the denominator is zero.
Undefined.
Rational expressions are fractions and are therefore undefined if the denominator is zero; the domain of a rational function is all real numbers except those that make the denominator of the related rational expression equal to 0. If a denominator contains variables, set it equal to zero and solve.
The domain of a rational function is the whole of the real numbers except those points where the denominator of the rational function, simplified if possible, is zero.
The answer depends on what w represents. If w is the denominator of the rational function then as w gets close to zero, the rational function tends toward plus or minus infinity - depending on the signs of the dominant terms in the numerator and denominator.
The Equation of a Rational Function has the Form,... f(x) = g(x)/h(x) where h(x) is not equal to zero. We will use a given Rational Function as an Example to graph showing the Vertical and Horizontal Asymptotes, and also the Hole in the Graph of that Function, if they exist. Let the Rational Function be,... f(x) = (x-2)/(x² - 5x + 6). f(x) = (x-2)/[(x-2)(x-3)]. Now if the Denominator (x-2)(x-3) = 0, then the Rational function will be Undefined, that is, the case of Division by Zero (0). So, in the Rational Function f(x) = (x-2)/[(x-2)(x-3)], we see that at x=2 or x=3, the Denominator is equal to Zero (0). But at x=3, we notice that the Numerator is equal to ( 1 ), that is, f(3) = 1/0, hence a Vertical Asymptote at x = 3. But at x=2, we have f(2) = 0/0, 'meaningless'. There is a Hole in the Graph at x = 2.
When the denominator is equal to zero, the expression is undefined. Close to those places, the expression tends towards plus infinity, or minus infinity. In other words, setting the denominator to zero will tell you where there are vertical asymptotes.
A zero of a function is a point at which the value of the function is zero. If you graph the function, it is a point at which the graph touches the x-axis.
A rational function is undefined - you might say that it "has a hole" - at any point where the denominator is zero. Assuming you mean "... hole at x = 2", any rational function which has the factor (x-2) in its denominator will have a hole at x = 2.
A rational function is the ratio of two polynomial functions. The function that is the denominator will have roots (or zeros) in the complex field and may have real roots. If it has real roots, then evaluating the rational function at such points will require division by zero. This is not defined. Since polynomials are continuous functions, their value will be close to zero near their roots. So, near a zero, the rational function will entail division by a very small quantity and this will result in the asymptotic behaviour.
True
It cannot be zero.