We set the denominator to zero to find the singularities: points where the graph is undefined.
A line which is the reflection of the original in y = x.
I am assuming the you are talking about the graph of the derivative. The graph of the derivative of F(x) is the graph such that, for any x, the value of x on the graph of the derivative of F(x) is the slope at point x in F(x).
a line graph
Bar graph
A linear graph contains both an x and y axis.
The answer depends on the form of the expression in the denominator. For example, the graph os 1/(1 + x2) has a pretty well-behaved graph, with a maximum vaue of 1 when x = 0 and asymptotes of y = 0
Simple answer, an x/y chart
We set the denominator to zero to find the singularities: points where the graph is undefined.
Discuss how you can use the zeros of the numerator and the zeros of the denominator of a rational function to determine whether the graph lies below or above the x-axis in a specified interval?
1 over surd x has no value since surd is not specific. In any case, you select a set of values for x from within the domain. Calculate the corresponding values of y and then plot each pair (x, y). Then, after checking that the denominator of the function does not become zero anywhere, you join the points by a smooth line. If the denominator does become zero, you select values for x close to where it does, calculate y and plot these as best you can. The graph will have a discontinuity where the denominator is zero.
Any graph in which there is at least one value of x for which there are more than one values of corresponding y. For example y = sqrt(x).
A line which is the reflection of the original in y = x.
I am assuming the you are talking about the graph of the derivative. The graph of the derivative of F(x) is the graph such that, for any x, the value of x on the graph of the derivative of F(x) is the slope at point x in F(x).
Any value of x which causes the denominator to equal zero. It's kind of vague, but if you mean the denominator to be (x raised to the 4th power), then x=0 must be excluded. If you mean (x + 4) then x=-4 will make the denominator equal zero, and if you mean (x-4) then x=4 will make it zero.
To translate the graph y = x to the graph of y = x - 6, shift the graph of y = x down 6 units.
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.