Question $y=3(2)^{2 x-8}+1$
D: $R:$ xint: yint: asymp: a) b)

Studdy Solution

STEP 1

What is this asking? We need to find the domain, range, x-intercept, y-intercept, and asymptote of the exponential function $y = 3(2)^{2x-8} + 1$ . Watch out! Remember, exponential functions have horizontal asymptotes, and we need to be careful when calculating the intercepts!

STEP 2

1. Define the function
2. Find the domain
3. Find the range
4. Find the x-intercept
5. Find the y-intercept
6. Find the asymptote

STEP 3

Alright, let's define our awesome function: $y = 3(2)^{2x-8} + 1$ .
This tells us how $y$ changes as $x$ changes.

STEP 4

The domain is all the possible $x$ values we can plug in.
Since we can plug any real number into the exponent, the domain is all real numbers!
We can write this as $(-\infty, \infty)$ .

STEP 5

The range is all the possible $y$ values that can come out.
Since $2$ raised to any power is always positive, the smallest value $(2)^{2x-8}$ can be is very close to zero, but never zero itself.

STEP 6

So, the smallest value of $3(2)^{2x-8}$ is also very close to zero, but never zero.
Adding one to this means the smallest value of $y$ is very close to one, but never one.
Therefore, the range is $(1, \infty)$ .

STEP 7

To find the x-intercept, we set $y = 0$ and solve for $x$ .
So, we have $0 = 3(2)^{2x-8} + 1$ .

STEP 8

Subtracting one from both sides gives $-1 = 3(2)^{2x-8}$ .
Dividing both sides by three gives $-\frac{1}{3} = (2)^{2x-8}$ .

STEP 9

Uh oh!
We can't raise $2$ to any power and get a negative number!
This means there's no x-intercept!
Our function never crosses the x-axis.

STEP 10

To find the y-intercept, we set $x = 0$ and solve for $y$ .
So, $y = 3(2)^{2(0)-8} + 1$ .

STEP 11

This simplifies to $y = 3(2)^{-8} + 1$ .
Since $2^{-8} = \frac{1}{2^8} = \frac{1}{256}$ , we have $y = 3 \cdot \frac{1}{256} + 1 = \frac{3}{256} + 1 = \frac{3+256}{256} = \frac{259}{256}$ .

STEP 12

So, the y-intercept is $\left(0, \frac{259}{256}\right)$ .

STEP 13

As $x$ becomes very negative, $(2)^{2x-8}$ gets very close to zero.
So, $3(2)^{2x-8}$ also gets very close to zero.
Adding one to this means $y$ gets very close to one.

STEP 14

Therefore, the horizontal asymptote is $y = 1$ .

STEP 15

Domain: $(-\infty, \infty)$ Range: $(1, \infty)$ x-intercept: None y-intercept: $\left(0, \frac{259}{256}\right)$ Asymptote: $y=1$

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