## 17Calculus Precalculus - Intercepts of Polynomials

The intercepts of a graph are the points where a graph crosses either the x-axis (x-intercepts) or the y-axis (y-intercepts). In calculus, we always work with real numbers unless explictly stated otherwise. So this page discusses only real x and y intercepts of polynomials. For intercepts of rational functions, see this page.

X-Intercepts

To determine the x-intercepts, we need to find where the graph crosses the x-axis. In these cases, the y-value is equal to zero. So in the equation, we set $$y=0$$ and then solve for the x-values. X-intercepts are also called zeros or roots. We discuss this in more detail on this page.

Y-Intercepts

To determine the y-intercepts, we need to find where the graph crosses the y-axis. In these cases, the x-value is equal to zero. So in the equation, we set $$x=0$$ and then solve for the y-values. If the equation represents a function (which all polynomials are), then there will be only one y-intercept since a function passes the vertical line test.

Practice

Unless other instructed, determine the intercepts (both x and y) for these functions.

$$y = 2x + 1$$

Problem Statement

Determine the x and y intercepts for $$y = 2x + 1$$.

x-intercept $$(-1/2, 0)$$
y-intercept $$(0,1)$$

Problem Statement

Determine the x and y intercepts for $$y = 2x + 1$$.

Solution

### 2786 video

x-intercept $$(-1/2, 0)$$
y-intercept $$(0,1)$$

$$y = x/2 - 3$$

Problem Statement

Determine the x and y intercepts for $$y = x/2 - 3$$.

x-intercept $$(6, 0)$$
y-intercept $$(0,-3)$$

Problem Statement

Determine the x and y intercepts for $$y = x/2 - 3$$.

Solution

### 2787 video

x-intercept $$(6, 0)$$
y-intercept $$(0,-3)$$

$$2x - 3y = 6$$

Problem Statement

Determine the x and y intercepts for $$2x - 3y = 6$$.

x-intercept $$(3, 0)$$
y-intercept $$(0,-2)$$

Problem Statement

Determine the x and y intercepts for $$2x - 3y = 6$$.

Solution

### 2788 video

x-intercept $$(3, 0)$$
y-intercept $$(0,-2)$$

$$3x + 4y = 12$$

Problem Statement

Determine the x and y intercepts for $$3x + 4y = 12$$.

x-intercept $$(4, 0)$$
y-intercept $$(0,3)$$

Problem Statement

Determine the x and y intercepts for $$3x + 4y = 12$$.

Solution

### 2789 video

x-intercept $$(4, 0)$$
y-intercept $$(0,3)$$

$$y = x^2 + 1$$

Problem Statement

Determine the intercepts (both x and y) for $$y = x^2 + 1$$.

x-intercepts: none
y-intercepts: $$y=1$$

Problem Statement

Determine the intercepts (both x and y) for $$y = x^2 + 1$$.

Solution

To find the x-intercepts, we set $$y=0$$ and solve the the x-values.
$$\begin{array}{rcl} 0 & = & x^2+1 \\ -1 & = & x^2 \end{array}$$
Since there are no real values that solve this equation, we know that there are no real x-intercepts.

Now let's determine the y-intercepts. To do this, we set $$x=0$$ and solve.
$$y=0^2+1 = 1$$
So, the only y-intercept is $$y=1$$.

x-intercepts: none
y-intercepts: $$y=1$$

$$y=2(x-1)(x+2)(x-3)$$

Problem Statement

Determine the intercepts (both x and y) for $$y=2(x-1)(x+2)(x-3)$$.

Solution

### 2532 video

video by PatrickJMT

$$y=x^2-3x-18$$

Problem Statement

Determine the intercepts (both x and y) for $$y=x^2-3x-18$$.

Solution

### 2533 video

video by PatrickJMT

$$f(x) = (x-2)(x+1)(x-4)(x+3)$$

Problem Statement

Determine the intercepts (both x and y) for $$f(x) = (x-2)(x+1)(x-4)(x+3)$$.

Solution

### 2534 video

video by MIP4U

$$f(x) = 2x^3 + 2x^2 - 24x$$

Problem Statement

Determine the intercepts (both x and y) for $$f(x) = 2x^3 + 2x^2 - 24x$$.

Solution

### 2535 video

$$y = -x^2 + 4x - 4$$

Problem Statement

Determine the intercepts (both x and y) for $$y = -x^2 + 4x - 4$$.

Solution

### 2557 video

Really UNDERSTAND Precalculus

### Calculus Topics Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

### Search Practice Problems

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem. The 17Calculus and 17Precalculus iOS and Android apps are no longer available for download. If you are still using a previously downloaded app, your app will be available until the end of 2020, after which the information may no longer be available. However, do not despair. All the information (and more) is now available on 17calculus.com for free.

You Can Have an Amazing Memory: Learn Life-Changing Techniques and Tips from the Memory Maestro Shop eBags.com, the leading online retailer of luggage, handbags, backpacks, accessories, and more! Try Amazon Music Unlimited Free Trial When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications.

DISCLAIMER - 17Calculus owners and contributors are not responsible for how the material, videos, practice problems, exams, links or anything on this site are used or how they affect the grades or projects of any individual or organization. We have worked, to the best of our ability, to ensure accurate and correct information on each page and solutions to practice problems and exams. However, we do not guarantee 100% accuracy. It is each individual's responsibility to verify correctness and to determine what different instructors and organizations expect. How each person chooses to use the material on this site is up to that person as well as the responsibility for how it impacts grades, projects and understanding of calculus, math or any other subject. In short, use this site wisely by questioning and verifying everything. If you see something that is incorrect, contact us right away so that we can correct it.