If you have dealt with algebraic equations, then you are probably familiar with the concept of intercepts. Intercepts are one of the fundamental components in algebra when graphing.
Intercepts are points on the graph of a function that cross either the x-axis or the y-axis. The x-intercept crosses the x-axis at the point (h,0). The y-intercept crosses the y-axis at the point (0,k).
I know this sounds simple, but let’s take a deeper dive into the definition of an intercept.
What are intercepts?
Intercepts are where the line or curve of a function crosses the x- or y-axis.
The point (or points) that we are interested in is the point that crosses the x-axis, or (h,0).
In most instances, to find the x-intercept, we can set the equation equal to zero a solve for x. Sometimes you will have to find the x-intercept for a line, and sometimes you will have to find the x-intercept for a polynomial. Each case will require a different approach, but they are all seeking to find where the function crosses or touches that x-axis.
The y-intercept is a bit easier to find. The point that we’re interest d in is (0,k). In this case, since x=0, we would replace all x’s with zeros and solve for y.
Let’s take a look at some examples. We’ll start with the most straightforward example, a line, and work towards a little harder ones.
How to find intercepts of a linear equation.
Two equations can be used to represent a line. A line can be represented by its slope-intercept form, y=ax+b, or by the standard form of Ax+By=C.
In this example, we’ll only touch on the slope-intercept form.
To find the x-intercepts of a line, we will have to set the equation equal to 0 and solve for x. We should get the point (h,0).
For instance, if we have the equation y=2x+5, we would set this equation equal to zero. When we set the equation equal to zero, we get 2x+5=0. When we solve for x, we get x=-5/2.
Our x-intercept for this equation is at (-5/2,0).
To find the y-intercept, we must set all x’s equal to 0 and solve for y. Remember, we are looking for the point that crosses the y-axis.
Using the same equation, y=2x+5, we replace all x’s in the equation with zeros.
We get y=0+5. When we solve for y, we get y=5. Our y-intercept is at (0,5).
It seems easy enough, right? But now, let’s try one a bit harder.
How to find the intercepts of a quadratic function.
A quadratic function is represented by the equation y=ax2+bx+c.
Let’s start with the y-intercept this time. The y-intercept of a quadratic function is located at the point (0,c). The y-intercept is the point where the x is equal to 0. When we replace all x’s in the function with zeros, we’re left with y=0+0+c.
Now let’s find t e x-intercept. The x-intercept occurs at the point where the y=0. Another way to say this is that we have to find the roots or the zeros of the function, which means we have to solve for x.
For a quadratic function, we have to factor the function down to quickly solve for x. If we can’t factor the equation, we’ll have to use the quadratic formula to find the roots or the zeros.
Let’s take a look at two examples. One will concentrate on factoring a quadratic function to find the roots, and for the other, we’ll have to use the quadratic formula to find the roots.
In this first example, we will work with the quadratic equation y=2x2+5x-3.
The first thing we will look for is the y-intercept since it’s the easiest to find. We set all x’s to zeros and solve for y. In this case, we get y=-3. Our y-intercept is located at (0,-3).
Then we move on to the x-intercepts. The easiest way to find them is to solve for x by finding the roots. The roots or solutions are what we get when we factor the equation and set each factor to zero.
For this equation, we can factor it into (2x-1) and (x+3). Once we have our factors, we can set each one equal to zero and solve for x.
Our x-intercepts are located at (1/2,0) and (-3,0) for this function.
Now let’s try on a bit harder. We’ll work with the quadratic equation y=8x2+43x+15.
We’ll find the y-intercepts first. We set all x’s equal to zeros and solve for y. In this case, we get y=15.
Our y-intercept is located at (0,15).
Now to find the x-intercepts.
Instead of working through trying to find the factors of this equation, we will use the quadratic formula instead.
Once we work through and substitute all values into the quadratic formula, we can find the x-intercepts. Our x-intercepts are at x=-5 and x=-6/16.
Our x-intercepts are located at (-5,0) and (-6/16,0).
The next equation we will work on is a rational function.
How to find the intercepts of a rational function.
A rational function consists of a ratio of polynomials. We need to take one crucial step before finding the x and y-intercepts.
We have to ensure that the rational function is reduced before we start looking for the x- or y-intercepts. This guarantees that we are not mistaking an intercept for a removable discontinuity.
The y-intercepts are found in the same way we found them for the other functions. We set all x’s to zero and solve for y. Easy-peasy.
The x-intercepts are a bit different. If we focus on setting the rational function to zero, we might end up with a zero in the denominator, and this would cause the whole function to be undefined. This is not what we want.
We have to focus on the numerator when looking for the x-intercepts. If the numerator is equal to zero, it sets the whole equation equal to zero.
This can only happen once we reduce the function and factor out any like terms in the numerator and denominator.
We’ll take a look at two examples. One will be a simple rational function, while the other will be a rational function that can be reduced.
The first example we’ll look at is y=(x-2)/(x-4).
This simple rational function is already in its reduced form since there are no common factors in the numerator and denominator.
The first thing we’ll look for is the y-intercept. We set all x’s equal to zero and solve for y. For this equation, we get y=(-4)/(-2). The y-intercept is located at the point (0,2)
Then we can move on to find the x-intercept. We don’t set the whole function to zero, just the numerator. For this equation, the numerator is x-4. When we set it equal to zero and solve for x, we get x=4.
Our x-intercept is located at the point (4,0).
The next example that we’ll look at is the equation y=(x2-3x-4)/(x2+4x+3).
Before we can find any intercept for this equation, we must check if there are any common factors in the numerator and denominator. We find the common factor of (x+1), and we cancel them out.
Once the function has been reduced, we can continue searching for the intercepts.
Once the function has been reduced, we can start by finding the y-intercept. We set all x’s to zero and solve for y. For this function, we have y=-4/3.
The y-intercept is located at the point (0,-4/3).
For the x-intercept, we only look at the numerator in the reduced function. We have x-4. We set this equal to zero and solve for x.
Our x-intercept for this function is located at the point (4,0).
Now let’s move on to an even trickier function, the polynomial.
How to find the intercepts of a polynomial.
Before we get started, I have to mention that we already touched on a few types of polynomials; a linear function and a quadratic function.
A polynomial is a function written with exponents or degrees that are whole numbers. The terms are written from the highest to the lowest degree in the standard polynomial form.
We have already had a preview of finding the intercepts of a polynomial. We’ll be taking those same concepts and using them on higher degree polynomials.
The y-intercept will always be the easiest intercept to find for a function. We replace all x’s with zeros and solve for y. No problem. We perform the same process no matter which function we have.
Finding the x-intercept of a polynomial function is the same as finding the x-intercept for a linear or quadratic equation.
In a polynomial function, the highest degree tells us about the possible number of x-intercepts. The highest degree is an indicator of how many x-intercepts we might be able to find for a function.
If we look at the linear example above, since the highest degree is one, we know there may be one possible x-intercept for our equation. In the quadratic function, the highest degree is two. We know that for this equation, there are two possible x-intercepts available for our function.
Let’s take a look at two example problems.
The first example is for the function y=x3+4x2+3x.
First, we look for the y-intercept. We replace all x’s with zeros, and we get y=0.
The y-intercept is located at the point (0,0).
Next, we find the x-intercept. Before we can find the x-intercept, we have to factor the equation. This equation can be factored into y=(x)(x+3)(x+1).
When we set each part equal to zero, we can find our x-intercepts for this function. We find the x-intercepts at the points (0,0), (-3,0), and (-1,0).
The last example we will look at is for the equation y=x3-1.
As always, we start with the y-intercept. Let’s replace all x’s with zeros and solve for y. In this case, we get y=-1.
The y-intercept is located at the point (0,-1).
The x-intercept can be found by either factoring this equation or setting the equation equal to zero and solving for x. I chose to go the easier route and set the equation equal to zero. I found that x3-1=0, and then I solved for x.
The x-intercept is located at the point (1,0).
I will not lie to you; math can be tricky sometimes. But taking the time to practice the homework and problems often will bring you the results you want.
Keep practicing, and don’t give up. 😄