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.
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You cannot, necessarily. Given a graph of the tan function, you could not.
No, a circle graph is never a function.
sine graph will be formed at origine of graph and cosine graph is find on y-axise
A line. The derivative of a function is its slope. If the slope is a constant then the graph is a line.
A derivative graph tracks the slope of a function.