Math / Algebra

QuestionGraph the given function by making a table of coordinates. $f(x)=\left(\frac{3}{4}\right)^{x}$
Complete the table of coordinates. \begin{tabular}{|c|c|c|c|c|c|} \hline $\mathbf{x}$ & -2 & -1 & 0 & 1 & 2 \\ \hline $\mathbf{y}$ & $\square$ & $\square$ & $\square$ & $\square$ & $\square$ \\ \hline \end{tabular} (Type integers or fractions. Simplify your answers.)

Studdy Solution

STEP 1

What is this asking? We need to fill in a table of $x$ and $y$ values for the function $f(x) = (\frac{3}{4})^x$ and then think about what the graph would look like! Watch out! Working with fractional bases can be a little tricky, so let's make sure we're careful with our calculations, especially when $x$ is negative!

STEP 2

1. Calculate $f(-2)$
2. Calculate $f(-1)$
3. Calculate $f(0)$
4. Calculate $f(1)$
5. Calculate $f(2)$

STEP 3

Let's start with $x = \mathbf{-2}$ .
We're going to substitute $-2$ into our function: $f(-2) = (\frac{3}{4})^{-2}$ .

STEP 4

Remember that a negative exponent means we flip the fraction and make the exponent positive.
So, $(\frac{3}{4})^{-2}$ becomes $(\frac{4}{3})^{2}$ .

STEP 5

Now, we just square the fraction: $(\frac{4}{3})^2 = \frac{4^2}{3^2} = \frac{\mathbf{16}}{\mathbf{9}}$ .
So, when $x = -2$ , $y = \frac{16}{9}$ .
Awesome!

STEP 6

Now for $x = \mathbf{-1}$ .
We have $f(-1) = (\frac{3}{4})^{-1}$ .

STEP 7

Again, flip that fraction to handle the negative exponent: $(\frac{3}{4})^{-1} = \frac{4}{3}$ .
So, when $x = -1$ , $y = \frac{\mathbf{4}}{\mathbf{3}}$ .

STEP 8

Time for $x = \mathbf{0}$ .
Anything raised to the power of 0 is 1 (except 0 itself, which we won't worry about here).
So, $f(0) = (\frac{3}{4})^0 = \mathbf{1}$ .

STEP 9

Let's plug in $x = \mathbf{1}$ .
This one is straightforward: $f(1) = (\frac{3}{4})^1 = \frac{\mathbf{3}}{\mathbf{4}}$ .

STEP 10

Finally, let's tackle $x = \mathbf{2}$ .
We have $f(2) = (\frac{3}{4})^2 = \frac{3^2}{4^2} = \frac{\mathbf{9}}{\mathbf{16}}$ .

STEP 11

Our completed table looks like this:
\begin{tabular}{|c|c|c|c|c|c|} \hline $x$ & -2 & -1 & 0 & 1 & 2 \\ \hline $y$ & $\frac{16}{9}$ & $\frac{4}{3}$ & 1 & $\frac{3}{4}$ & $\frac{9}{16}$ \\ \hline \end{tabular}
If we were to graph these points, we'd see a nice, smooth curve decreasing from left to right!

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Studdy Solution

STEP 1

STEP 2

1. Calculate $f(-2)$
2. Calculate $f(-1)$
3. Calculate $f(0)$
4. Calculate $f(1)$
5. Calculate $f(2)$

STEP 3

Let's start with $x = \mathbf{-2}$ .
We're going to substitute $-2$ into our function: $f(-2) = (\frac{3}{4})^{-2}$ .

STEP 4

Remember that a negative exponent means we flip the fraction and make the exponent positive.
So, $(\frac{3}{4})^{-2}$ becomes $(\frac{4}{3})^{2}$ .

STEP 5

Now, we just square the fraction: $(\frac{4}{3})^2 = \frac{4^2}{3^2} = \frac{\mathbf{16}}{\mathbf{9}}$ .
So, when $x = -2$ , $y = \frac{16}{9}$ .
Awesome!

STEP 6

Now for $x = \mathbf{-1}$ .
We have $f(-1) = (\frac{3}{4})^{-1}$ .

STEP 7

Again, flip that fraction to handle the negative exponent: $(\frac{3}{4})^{-1} = \frac{4}{3}$ .
So, when $x = -1$ , $y = \frac{\mathbf{4}}{\mathbf{3}}$ .

STEP 8

Time for $x = \mathbf{0}$ .
Anything raised to the power of 0 is 1 (except 0 itself, which we won't worry about here).
So, $f(0) = (\frac{3}{4})^0 = \mathbf{1}$ .

STEP 9

Let's plug in $x = \mathbf{1}$ .
This one is straightforward: $f(1) = (\frac{3}{4})^1 = \frac{\mathbf{3}}{\mathbf{4}}$ .

STEP 10

Finally, let's tackle $x = \mathbf{2}$ .
We have $f(2) = (\frac{3}{4})^2 = \frac{3^2}{4^2} = \frac{\mathbf{9}}{\mathbf{16}}$ .

STEP 11

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Studdy Solution

STEP 1

STEP 2

1. Calculate f(−2)f(-2)f(−2)2. Calculate f(−1)f(-1)f(−1)3. Calculate f(0)f(0)f(0)4. Calculate f(1)f(1)f(1)5. Calculate f(2)f(2)f(2)

STEP 3

Let's **start** with x=−2x = \mathbf{-2}x=−2.We're going to substitute −2-2−2 into our function: f(−2)=(34)−2f(-2) = (\frac{3}{4})^{-2}f(−2)=(43​)−2.

STEP 4

Remember that a negative exponent means we **flip** the fraction and make the exponent positive.So, (34)−2(\frac{3}{4})^{-2}(43​)−2 becomes (43)2(\frac{4}{3})^{2}(34​)2.

STEP 5

Now, we just **square** the fraction: (43)2=4232=169(\frac{4}{3})^2 = \frac{4^2}{3^2} = \frac{\mathbf{16}}{\mathbf{9}}(34​)2=3242​=916​.So, when x=−2x = -2x=−2, y=169y = \frac{16}{9}y=916​.Awesome!

STEP 6

Now for x=−1x = \mathbf{-1}x=−1.We have f(−1)=(34)−1f(-1) = (\frac{3}{4})^{-1}f(−1)=(43​)−1.

STEP 7

Again, **flip** that fraction to handle the negative exponent: (34)−1=43(\frac{3}{4})^{-1} = \frac{4}{3}(43​)−1=34​.So, when x=−1x = -1x=−1, y=43y = \frac{\mathbf{4}}{\mathbf{3}}y=34​.

STEP 8

Time for x=0x = \mathbf{0}x=0.Anything raised to the power of 0 is 1 (except 0 itself, which we won't worry about here).So, f(0)=(34)0=1f(0) = (\frac{3}{4})^0 = \mathbf{1}f(0)=(43​)0=1.

STEP 9

Let's **plug in** x=1x = \mathbf{1}x=1.This one is straightforward: f(1)=(34)1=34f(1) = (\frac{3}{4})^1 = \frac{\mathbf{3}}{\mathbf{4}}f(1)=(43​)1=43​.

STEP 10

Finally, let's **tackle** x=2x = \mathbf{2}x=2.We have f(2)=(34)2=3242=916f(2) = (\frac{3}{4})^2 = \frac{3^2}{4^2} = \frac{\mathbf{9}}{\mathbf{16}}f(2)=(43​)2=4232​=169​.

STEP 11

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1. Calculate $f(-2)$
2. Calculate $f(-1)$
3. Calculate $f(0)$
4. Calculate $f(1)$
5. Calculate $f(2)$

Let's start with $x = \mathbf{-2}$ .
We're going to substitute $-2$ into our function: $f(-2) = (\frac{3}{4})^{-2}$ .

Remember that a negative exponent means we flip the fraction and make the exponent positive.
So, $(\frac{3}{4})^{-2}$ becomes $(\frac{4}{3})^{2}$ .

Now, we just square the fraction: $(\frac{4}{3})^2 = \frac{4^2}{3^2} = \frac{\mathbf{16}}{\mathbf{9}}$ .
So, when $x = -2$ , $y = \frac{16}{9}$ .
Awesome!

Now for $x = \mathbf{-1}$ .
We have $f(-1) = (\frac{3}{4})^{-1}$ .

Again, flip that fraction to handle the negative exponent: $(\frac{3}{4})^{-1} = \frac{4}{3}$ .
So, when $x = -1$ , $y = \frac{\mathbf{4}}{\mathbf{3}}$ .

Time for $x = \mathbf{0}$ .
Anything raised to the power of 0 is 1 (except 0 itself, which we won't worry about here).
So, $f(0) = (\frac{3}{4})^0 = \mathbf{1}$ .

Let's plug in $x = \mathbf{1}$ .
This one is straightforward: $f(1) = (\frac{3}{4})^1 = \frac{\mathbf{3}}{\mathbf{4}}$ .

Finally, let's tackle $x = \mathbf{2}$ .
We have $f(2) = (\frac{3}{4})^2 = \frac{3^2}{4^2} = \frac{\mathbf{9}}{\mathbf{16}}$ .