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Question

2
1
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x+3
−2, following the instructions below.
Equation of asymptote:
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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.