Analyzing F(x)=(x^2-5x+6)/(x^2-10x+16) Domain Intercepts And Asymptotes

by Chloe Fitzgerald 72 views

Hey there, math enthusiasts! Today, we're diving deep into the fascinating world of rational functions, and we're going to dissect the function $f(x)=\frac{x2-5x+6}{x2-10x+16}$. Buckle up, because we're about to explore its domain, intercepts, holes, and asymptotes. By the end of this journey, you'll have a solid understanding of how to analyze and graph such functions. So, let’s get started, guys!

1. Domain: Where Does Our Function Live?

Let's kick things off by figuring out the domain of our function. Remember, the domain is the set of all possible input values (x-values) for which the function is defined. In simpler terms, it's where our function is happy and produces a real output. For rational functions like ours, the main thing that can make the function unhappy is division by zero. We need to identify those pesky x-values that make the denominator equal to zero and exclude them from the domain. After all, dividing by zero is a big no-no in the math world!

So, let's focus on the denominator: $x^2-10x+16$. To find the values that make it zero, we need to solve the equation $x^2-10x+16 = 0$. This looks like a quadratic equation, and we can solve it by factoring. Can you guys spot the factors? We're looking for two numbers that multiply to 16 and add up to -10. Those numbers are -2 and -8. So, we can factor the quadratic as follows:

(xβˆ’2)(xβˆ’8)=0(x-2)(x-8) = 0

Now, we can use the zero-product property, which tells us that if the product of two factors is zero, then at least one of the factors must be zero. This gives us two possible solutions:

  • xβˆ’2=0β‡’x=2x - 2 = 0 \Rightarrow x = 2

  • xβˆ’8=0β‡’x=8x - 8 = 0 \Rightarrow x = 8

These are the x-values that make the denominator zero, and we must exclude them from the domain. Therefore, the domain of our function is all real numbers except 2 and 8. To express this in interval notation, we write:

(βˆ’βˆž,2)βˆͺ(2,8)βˆͺ(8,∞)(-\infty, 2) \cup (2, 8) \cup (8, \infty)

This notation means that the domain includes all numbers from negative infinity up to 2 (but not including 2), then all numbers between 2 and 8 (but not including 2 and 8), and finally all numbers from 8 to positive infinity (but not including 8). Essentially, we've created a map of the x-values where our function is valid and well-behaved.

2. Y-Intercept(s): Where Does Our Function Cross the Y-Axis?

Next up, let's find the y-intercept(s) of our function. The y-intercept is the point(s) where the graph of the function crosses the y-axis. This happens when x is equal to 0. To find the y-intercept, we simply substitute x = 0 into our function and see what y-value we get. It's like asking our function: "Hey, where do you hang out on the y-axis?"

So, let's plug in x = 0 into our function:

f(0)=02βˆ’5(0)+602βˆ’10(0)+16=616=38f(0) = \frac{0^2 - 5(0) + 6}{0^2 - 10(0) + 16} = \frac{6}{16} = \frac{3}{8}

This tells us that the function intersects the y-axis at y = 3/8. Therefore, the y-intercept is the point (0, 3/8). This is a key point on our graph, giving us a crucial anchor for our function's behavior.

3. X-Intercept(s): Where Does Our Function Cross the X-Axis?

Now, let's hunt for the x-intercept(s). The x-intercept(s) are the point(s) where the graph of the function crosses the x-axis. This occurs when y (or f(x)) is equal to 0. To find the x-intercept(s), we need to solve the equation f(x) = 0. Remember, a fraction is equal to zero only when its numerator is zero. So, we can focus on the numerator of our function. Essentially, we're asking: "Hey function, when do you hit zero?"

The numerator of our function is $x^2 - 5x + 6$. Let's set it equal to zero and solve:

x2βˆ’5x+6=0x^2 - 5x + 6 = 0

This is another quadratic equation. Let's try factoring it. We need two numbers that multiply to 6 and add up to -5. Those numbers are -2 and -3. So, we can factor the quadratic as:

(xβˆ’2)(xβˆ’3)=0(x - 2)(x - 3) = 0

Using the zero-product property again, we get two possible solutions:

  • xβˆ’2=0β‡’x=2x - 2 = 0 \Rightarrow x = 2

  • xβˆ’3=0β‡’x=3x - 3 = 0 \Rightarrow x = 3

Wait a minute! We found that x = 2 is a solution, but we already know from our domain analysis that x = 2 is not in the domain. This means that x = 2 is not an x-intercept. It's something else entirely, which we'll discuss in the next section (hint: it's a hole!).

So, the only x-intercept we have is x = 3. This means the function crosses the x-axis at the point (3, 0). This point, along with our y-intercept, gives us a good sense of how our function is positioned on the coordinate plane.

4. X-Value of Any Holes: Unmasking the Invisible Gaps

Ah, holes! These are the sneaky little gaps in the graph of our function. They occur when a factor cancels out from both the numerator and the denominator. Remember that x = 2 was a solution to the numerator, but it was also a value excluded from the domain. This is a classic sign of a hole. Holes are like secret passages in the function's landscape. They represent points where the function is undefined, but the graph appears to be continuous if you didn't know any better.

To find the x-value of the hole, we already did the hard work. It's the value that makes both the numerator and denominator zero, which is x = 2. This is a crucial piece of information. We know our function has a discontinuity at x=2, but it's not a vertical asymptote.

To find the y-value of the hole, we need to "plug" the x-value (x = 2) into the simplified form of the function (after canceling out the common factor). First, let's factor both the numerator and the denominator:

f(x)=x2βˆ’5x+6x2βˆ’10x+16=(xβˆ’2)(xβˆ’3)(xβˆ’2)(xβˆ’8)f(x) = \frac{x^2 - 5x + 6}{x^2 - 10x + 16} = \frac{(x - 2)(x - 3)}{(x - 2)(x - 8)}

Now, we can cancel out the common factor (x - 2), but remember, this cancellation is only valid if x is not equal to 2. This gives us the simplified form:

f(x)=xβˆ’3xβˆ’8,xβ‰ 2f(x) = \frac{x - 3}{x - 8}, \quad x \neq 2

Now, let's plug in x = 2 into the simplified form:

f(2)=2βˆ’32βˆ’8=βˆ’1βˆ’6=16f(2) = \frac{2 - 3}{2 - 8} = \frac{-1}{-6} = \frac{1}{6}

So, the hole is located at the point (2, 1/6). This means that the graph of our function has a tiny little gap at this point, like a missing stitch in a beautiful tapestry.

5. Vertical Asymptotes: The Walls Our Function Can't Cross

Finally, let's talk about vertical asymptotes. These are the vertical lines that the graph of the function approaches but never quite touches. Vertical asymptotes occur at the x-values that make the denominator of the simplified function equal to zero. They act like walls, guiding the function's behavior as it gets closer and closer to these lines.

We already simplified our function to:

f(x)=xβˆ’3xβˆ’8,xβ‰ 2f(x) = \frac{x - 3}{x - 8}, \quad x \neq 2

The denominator of the simplified function is (x - 8). Setting it equal to zero, we get:

xβˆ’8=0β‡’x=8x - 8 = 0 \Rightarrow x = 8

So, there is a vertical asymptote at x = 8. This means that as x approaches 8 from either the left or the right, the function's value will shoot off towards positive or negative infinity.

Conclusion: Putting It All Together

Wow, we've covered a lot! We've successfully dissected the function $f(x)=\frac{x2-5x+6}{x2-10x+16}$ and uncovered its secrets. We found its domain, y-intercept, x-intercept, hole, and vertical asymptote. This information allows us to sketch a pretty accurate graph of the function and understand its behavior. By finding these key features, we can truly understand the behavior of this rational function.

Remember, analyzing rational functions is like solving a puzzle. Each piece of information – the domain, intercepts, holes, and asymptotes – contributes to the overall picture. So, keep practicing, and you'll become a master of rational functions in no time! I hope this guide has been helpful, guys! Keep exploring the amazing world of mathematics!