How To Find X-Intercepts Of A Rational Function

Hey there, math enthusiasts! Today, we're diving deep into the world of rational functions and tackling a common, yet sometimes tricky, problem: finding those elusive x-intercepts. You know, those points where the graph of our function decides to say hello to the x-axis. It's a super important concept, especially when you're analyzing graphs or solving equations. So, grab your calculators, your notebooks, and maybe a comfy seat, because we're about to break down how to find the x-intercepts for the function $f(x)=\frac{2 x^2-14 x}{3 x^2-28 x-20}$. We'll be showing you step-by-step how to solve this, so don't worry if you're feeling a little rusty on the algebra. Our goal here is to make this whole process crystal clear for you, guys, so you can conquer any similar problems that come your way. We'll cover what x-intercepts are, why they matter, and then get straight into the nitty-gritty of solving this specific function. Let's get started!

Understanding X-Intercepts

Alright, so before we jump into solving our specific function, let's get on the same page about what x-intercepts actually are. Think of the x-axis as the horizontal line where the y-value is always zero. So, an x-intercept is simply a point on the graph of a function where the function's value, or the y-value, is equal to zero. In other words, it's where the graph crosses or touches the x-axis. Mathematically, this means we're looking for the values of $x$ that make $f(x) = 0$. Why is this so crucial? Well, x-intercepts give us vital information about the behavior of a function. They tell us the roots or solutions of the equation $f(x) = 0$, which are fundamental to understanding where a function is positive, negative, or equal to zero. For polynomial functions, the number of x-intercepts can tell us about the number of real roots. For rational functions, like the one we're about to tackle, finding x-intercepts involves a bit of a twist because we're dealing with both a numerator and a denominator. The general idea is still the same: set the function equal to zero and solve for $x$. However, we must always be mindful of the denominator. If a value of $x$ makes the denominator zero, it's not a valid input for the function, and therefore it can't be an x-intercept. So, when we're dealing with rational functions, the x-intercepts occur when the numerator is equal to zero AND the denominator is NOT equal to zero for that same value of $x$. This distinction is super important, and we'll see why it matters in our example. Understanding this core concept is the first big step to acing these problems. It’s all about where the function meets the x-axis, and for rational functions, we just have to be extra careful about the bottom part of the fraction.

Solving for X-Intercepts: Step-by-Step

Now that we've got a solid grasp on what x-intercepts are, let's roll up our sleeves and find them for our function: $f(x)=\frac{2 x^2-14 x}{3 x^2-28 x-20}$. Remember our rule: for a rational function, x-intercepts happen when the numerator equals zero AND the denominator does NOT equal zero. So, the first order of business is to set the numerator equal to zero and solve for $x$. Our numerator is $2x^2 - 14x$. Let's set it equal to zero:

$2x^2 - 14x = 0$

To solve this quadratic equation, we can factor out the greatest common factor, which is $2x$. So, we get:

$2x(x - 7) = 0$

For this product to be zero, at least one of the factors must be zero. This gives us two potential solutions:

1. $2x = 0 \implies x = 0$
2. $x - 7 = 0 \implies x = 7$

So, we have two potential candidates for our x-intercepts: $x = 0$ and $x = 7$. But here's the crucial part, guys: we must check if these values of $x$ make the denominator zero. If they do, then they are not x-intercepts. Our denominator is $3x^2 - 28x - 20$. Let's test our potential x-values.

Checking the Denominator for $x = 0$

Let's substitute $x = 0$ into the denominator:

$3(0)^2 - 28(0) - 20 = 0 - 0 - 20 = -20$

Since the denominator is $-20$ (which is definitely not zero) when $x = 0$, this value of $x$ is a valid x-intercept. So, one of our x-intercepts is at the coordinate point (0, 0). Pretty straightforward so far, right?

Checking the Denominator for $x = 7$

Now, let's substitute $x = 7$ into the denominator:

$3(7)^2 - 28(7) - 20$

First, calculate $7^2 = 49$:

$3(49) - 28(7) - 20$

Now, perform the multiplications:

$3(49) = 147$ $28(7) = 196$

So, the expression becomes:

$147 - 196 - 20$

Let's do the subtraction:

$147 - 196 = -49$

And then:

$-49 - 20 = -69$

Since the denominator is $-69$ (again, not zero) when $x = 7$, this value of $x$ is also a valid x-intercept. So, another x-intercept is at the coordinate point (7, 0).

What About the Denominator's Roots?

Okay, so we've found our x-intercepts by setting the numerator to zero and checking that the denominator isn't zero for those specific x-values. But what if the denominator itself has roots? What does that mean for our function? It means those values of $x$ are undefined for our function, and they typically correspond to vertical asymptotes or holes in the graph. It's super important to find these as well to get a complete picture of the function's behavior, even though they aren't x-intercepts. Let's find the roots of our denominator, $3x^2 - 28x - 20 = 0$. This is a quadratic equation, and we can solve it using the quadratic formula or by factoring. Let's try factoring first. We're looking for two numbers that multiply to $(3)(-20) = -60$ and add up to $-28$. After a bit of trial and error (or if you're good at spotting these!), you might find that $-30$ and $2$ fit the bill: $(-30)(2) = -60$ and $-30 + 2 = -28$. Now we can rewrite the middle term:

$3x^2 - 30x + 2x - 20 = 0$

Now, we factor by grouping:

$3x(x - 10) + 2(x - 10) = 0$

Factor out the common term $(x - 10)$:

$(3x + 2)(x - 10) = 0$

This gives us two potential values that make the denominator zero:

1. $3x + 2 = 0 \implies 3x = -2 \implies x = -\frac{2}{3}$
2. $x - 10 = 0 \implies x = 10$

So, $x = -\frac{2}{3}$ and $x = 10$ are the values where our function is undefined. Since neither of these values made our numerator zero (remember, our numerator's roots are 0 and 7), these will be vertical asymptotes for the graph of $f(x)$. This confirms that our x-intercepts ($x=0$ and $x=7$) are valid because they don't coincide with the values that make the denominator zero. It's this careful checking process that ensures accuracy, guys!

Final Answer: The X-Intercepts

After carefully going through all the steps, we've determined the x-intercepts of the function $f(x)=\frac{2 x^2-14 x}{3 x^2-28 x-20}$. We found that the numerator, $2x^2 - 14x$, equals zero when $x=0$ and $x=7$. We then verified that neither of these values makes the denominator, $3x^2 - 28x - 20$, equal to zero. The denominator is $-20$ when $x=0$ and $-69$ when $x=7$. Because these conditions are met, both $x=0$ and $x=7$ correspond to x-intercepts. It's crucial to remember that for rational functions, you must always check the denominator! The x-intercepts are the points where the graph crosses the x-axis, and they are always written as coordinate pairs $(x, y)$. Since the y-value at an x-intercept is always 0, our intercepts are:

• (0, 0)
• (7, 0)

So, there you have it! Two x-intercepts for our function. Keep practicing these steps, and you'll be a pro at finding x-intercepts in no time. Remember to always check your denominator!