Question

Use technology to find points and then graph the function y, equals, left parenthesis, one half, right parenthesis, start superscript, x, plus, 3, end superscript, minus, 2, commay=(
2
1
​
)
x+3
−2, following the instructions below.
Equation of asymptote:
y, equalsy=

Plot at least four points with integer coordinates

Answer 1

To analyze the function \( y = \left( \frac{1}{2} \right)^{(x + 3)} - 2 \) and find its properties, we can first determine the equation of the asymptote and then find integer coordinate points for plotting.

### Equation of the Asymptote

For exponential functions of the form \( y = a \cdot b^{(x + h)} + k \), the horizontal asymptote is given by \( y = k \).

In our function, we have:

- \( a = 1 \)
- \( b = \frac{1}{2} \)
- \( h = 3 \)
- \( k = -2 \)

Therefore, the equation of the horizontal asymptote is:
\[
y = -2
\]

### Finding Points with Integer Coordinates

We will now calculate the value of \( y \) for several integer values of \( x \) and identify at least four points with integer coordinates.

Let's choose integer values of \( x \) around the vertical shift:

1. **For \( x = -5 \)**:
\[
y = \left( \frac{1}{2} \right)^{-5 + 3} - 2 = \left( \frac{1}{2} \right)^{-2} - 2 = 4 - 2 = 2
\]
Point: \((-5, 2)\)

2. **For \( x = -4 \)**:
\[
y = \left( \frac{1}{2} \right)^{-4 + 3} - 2 = \left( \frac{1}{2} \right)^{-1} - 2 = 2 - 2 = 0
\]
Point: \((-4, 0)\)

3. **For \( x = -3 \)**:
\[
y = \left( \frac{1}{2} \right)^{-3 + 3} - 2 = \left( \frac{1}{2} \right)^{0} - 2 = 1 - 2 = -1
\]
Point: \((-3, -1)\)

4. **For \( x = -2 \)**:
\[
y = \left( \frac{1}{2} \right)^{-2 + 3} - 2 = \left( \frac{1}{2} \right)^{1} - 2 = \frac{1}{2} - 2 = -\frac{3}{2}
\]
Point: \((-2, -\frac{3}{2})\) (This is not an integer point, so we will skip this.)

5. **For \( x = -1 \)**:
\[
y = \left( \frac{1}{2} \right)^{-1 + 3} - 2 = \left( \frac{1}{2} \right)^{2} - 2 = \frac{1}{4} - 2 = -\frac{7}{4}
\]
Point: \((-1, -\frac{7}{4})\)

6. **For \( x = 0 \)**:
\[
y = \left( \frac{1}{2} \right)^{0 + 3} - 2 = \left( \frac{1}{2} \right)^{3} - 2 = \frac{1}{8} - 2 = -\frac{15}{8}
\]
Point: \((0, -\frac{15}{8})\)

7. **For \( x = 1 \)**:
\[
y = \left( \frac{1}{2} \right)^{1 + 3} - 2 = \left( \frac{1}{2} \right)^{4} - 2 = \frac{1}{16} - 2 = -\frac{31}{16}
\]
Point: \((1, -\frac{31}{16})\)

8. **For \( x = 2 \)**:
\[
y = \left( \frac{1}{2} \right)^{2 + 3} - 2 = \left( \frac{1}{2} \right)^{5} - 2 = \frac{1}{32} - 2 = -\frac{63}{32}
\]
Point: \((2, -\frac{63}{32})\)

Based on our calculations, we can summarize the integer coordinate points we found:

- \((-5, 2)\)
- \((-4, 0)\)
- \((-3, -1)\)

As additional points with integer values on \(y\):

- Any additional integer \(x\) may yield non-integers or fractional values.

### Summary of Points:

1. \((-5, 2)\)
2. \((-4, 0)\)
3. \((-3, -1)\)

### Graphing

You can plot these points on a Cartesian plane. The horizontal asymptote is \(y = -2\). The curve derived from this function will approach this asymptote as \(x\) goes towards positive or negative infinity.