How to find x intercept of a rational function

Updated April 24, 2017

By Karl Wallulis

The intercepts of a function are the values of x when f(x) = 0 and the value of f(x) when x = 0, corresponding to the coordinate values of x and y where the graph of the function crosses the x- and y-axes. Find the y-intercept of a rational function as you would for any other type of function: plug in x = 0 and solve. Find the x-intercepts by factoring the numerator. Remember to exclude holes and vertical asymptotes when finding the intercepts.

Plug the value x = 0 into the rational function and determine the value of f(x) to find the y-intercept of the function. For example, plug x = 0 into the rational function f(x) = (x^2 - 3x + 2) / (x - 1) to get the value (0 - 0 + 2) / (0 - 1), which is equal to 2 / -1 or -2 (if the denominator is 0, there is a vertical asymptote or hole at x = 0 and therefore no y-intercept). The y-intercept of the function is y = -2.

Factor the numerator of the rational function completely. In the above example, factor the expression (x^2 - 3x + 2) into (x - 2)(x - 1).

Set the factors of the numerator equal to 0 and solve for the value of the variable to find the potential x-intercepts of the rational function. In the example, set the factors (x - 2) and (x - 1) equal to 0 to get the values x = 2 and x = 1.

Plug the values of x you found in Step 3 into the rational function to verify that they are x-intercepts. X-intercepts are values of x that make the function equal to 0. Plug x = 2 into the example function to get (2^2 - 6 + 2) / (2 - 1), which equals 0 / -1 or 0, so x = 2 is an x-intercept. Plug x = 1 into the function to get (1^2 - 3 + 2) / (1 - 1) to get 0 / 0, which means there is a hole at x = 1, so there is only one x-intercept, x = 2.

In this tutorial the instructor shows how to find the x and y intercepts of rational functions. Finding the intercepts of a rational function is similar to finding the intercepts of other normal equations. You can find the x intercept of the equation by setting the value of y to zero and solving the equation. Similarly you can solve the y intercept by setting the value of x to zero and solving the equation. Now while solving this rational function for intercepts if you face a situation where the value in the denominator is zero it implies that there is no intercept in that case as dividing something by zero is indeterminate. This video shows how to solve x and y intercepts of rational functions.

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The $y$ -intercept of a function is the $y$ -coordinate of the point where the function crosses the $y$ -axis.

The value of the $y$ -intercept of $y = f (x)$ is numerically equal to $f (0)$ . A function can have at most one $y$ -intercept, as it can have at most one value of $f (0)$ .

The $y$ -intercept of the graph can be obtained by setting $x = 0$ , and thus we get $\times 0 + 4 = 4.$ $□_\square$

The $y$ -intercept can be obtained by setting $x = 0$ , and thus we get $y = e^{0} + 4 = 1+ 4 = 5$ . $□_\square$

The $y$ -intercept can be obtained by setting $x = 0$ , and thus we get $\dfrac{9}{3} = 3$ . $□_\square$

Setting $x = 0$ , we get $y =0^3 + 3×0^2 + 3×0 + 3 = 3$ . $□_\square$

The $x$ -intercept of a function is the $x$ -coordinate of the point where the function crosses the $x$ -axis.

Since the function can cross the $x$ -axis multiple times, it can have multiple $x$ -intercepts. The values of $x$ -intercepts of $y = f (x)$ can be obtained by setting $f (x) = 0$ .

The $x$ -intercept can be obtained by setting $y = 0$ , so we get $\implies x = -\dfrac{5}{2}$ . $□_\square$

We can get the $x$ -intercept by setting $y = 0$ , so we get $x^2 - 5x + 6= 0$ .
It can be factored as $(x - 2) (x - 3) = 0$ , which implies $x = 2, 3$ .
Thus, the $x$ -intercepts of the graph are $2$ and $3$ , so the sum of the $x$ -intercepts is $2 + 3 = 5$ . $□_\square$

We can get the $x$ -intercepts by setting $y = 0$ , so we get $e^x = -1$ .

However, $e^x$ is a positive quantity for all real values of $x$ and can never be equal to $- 1$ . Hence, this equation has no solutions, and thus the number of $x$ -intercepts in the graph is 0. $□_\square$

We can get the $x$ intercepts by setting $\dfrac{(x-1)(x-2)(x-3)}{x(x-1)(x+1)} = 0$ .
Hence $\implies x = 1, 2, 3$ .
However, $\neq -1, 0, 1$ because each of these values of $x$ makes the denominator zero.
So the $x$ -intercepts are $x = 2, 3$ , and thus their product is $\times 3 = 6$ . $□_\square$