Math  /  Algebra

Questionf(x)=x2162x216x+32f(x)=\frac{x^{2}-16}{2 x^{2}-16 x+32}
Answer Attempt 1 out of 2
Horizontal Asymptote: y=y= \square No horizontal asymptote \square No vertical asymptote
Vertical Asymptote: x=x= No vertical asymptote xx-Intercept: \square ,0) \square yy-Intercept: (0, \square No yy-intercept \square Hole: \square \square No hole

Studdy Solution

STEP 1

What is this asking? We're exploring a function to find its horizontal and vertical asymptotes, x and y intercepts, and any holes it might have! Watch out! Don't forget to **factor** everything first.
It makes life so much easier!
Also, remember a **hole** appears when a factor cancels out in both the numerator and denominator.

STEP 2

1. Simplify the Function
2. Horizontal Asymptote
3. Vertical Asymptote
4. x-intercept
5. y-intercept
6. Hole

STEP 3

Alright, let's **factor** that function!
We can rewrite the numerator x216x^2 - 16 as (x4)(x+4)(x-4) \cdot (x+4) – that's a difference of squares!
And the denominator, 2x216x+322x^2 - 16x + 32, can be factored as 2(x28x+16)2 \cdot (x^2 - 8x + 16), which further simplifies to 2(x4)(x4)2 \cdot (x-4) \cdot (x-4) or 2(x4)22(x-4)^2.

STEP 4

So, our simplified function looks like this: f(x)=(x4)(x+4)2(x4)(x4) f(x) = \frac{(x-4)(x+4)}{2(x-4)(x-4)} We can divide to one the term (x4)(x-4), giving us: f(x)=x+42(x4), for x4 f(x) = \frac{x+4}{2(x-4)}, \text{ for } x \neq 4 Hold on to that x4x \neq 4 piece of information, it's important for later!

STEP 5

For the **horizontal asymptote**, we look at what happens when xx gets REALLY big.
Since the **highest powers** of xx are the same in the numerator and denominator (both x1x^1), we divide the **leading coefficients**.
The numerator's leading coefficient is **1** and the denominator's is **2**.

STEP 6

So, our horizontal asymptote is y=12 y = \frac{1}{2} .
Boom!

STEP 7

**Vertical asymptotes** happen when the denominator is zero and the numerator isn't.
Looking at our simplified function, the denominator 2(x4)2(x-4) is zero when x=4x = 4.

STEP 8

However, remember that x4x \neq 4 from when we simplified the function?
That means there's a hole at x=4x=4, not a vertical asymptote.
So, there's *no* vertical asymptote!

STEP 9

The **x-intercept** occurs when f(x)=0f(x) = 0, which means the numerator must be zero.
In our simplified function, the numerator x+4x+4 is zero when x=4x = -4.

STEP 10

So, our x-intercept is (4,0)(-4, 0).
Nice!

STEP 11

The **y-intercept** happens when x=0x = 0.
Let's plug that into our simplified function: f(0)=0+42(04)=42(4)=48=12 f(0) = \frac{0+4}{2(0-4)} = \frac{4}{2(-4)} = \frac{4}{-8} = -\frac{1}{2}

STEP 12

Therefore, our y-intercept is (0,12)(0, -\frac{1}{2}).
Fantastic!

STEP 13

Remember how we divided to one the (x4)(x-4) term earlier?
That means there's a **hole** at x=4x=4.
To find the y-coordinate of the hole, we plug x=4x=4 into the simplified function: 4+42(44) \frac{4+4}{2(4-4)} Uh oh!
The denominator is zero, which confirms there's a hole.

STEP 14

To find the y-coordinate, we use the simplified version, x+42(x4) \frac{x+4}{2(x-4)} , and plug in x=4x=4: limx4x+42(x4) \lim_{x \to 4} \frac{x+4}{2(x-4)} We can't directly substitute x=4x=4 because it leads to division by zero.
However, we can observe that as xx approaches 4, the expression approaches 80\frac{8}{0}, which indicates a vertical asymptote or a hole.
Since we've already established there's no vertical asymptote, it must be a hole at x=4x=4.

STEP 15

Let's evaluate the limit as xx approaches 4: limx4x+42(x4) \lim_{x \to 4} \frac{x+4}{2(x-4)} As xx approaches 4 from the left (values slightly less than 4), the expression approaches -\infty.
As xx approaches 4 from the right (values slightly greater than 4), the expression approaches ++\infty.
This confirms the presence of a hole at x=4x=4.

STEP 16

Horizontal Asymptote: y=12y = \frac{1}{2} Vertical Asymptote: No vertical asymptote xx-Intercept: (4,0)(-4, 0) yy-Intercept: (0,12)(0, -\frac{1}{2}) Hole: x=4x = 4

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