Question

use technology to find points and then graph the function $y = 2^{x - 2} - 5$, following the instructions below.
equation of asymptote:
$y = -5$
plot at least four points with integer coordinates that fit on the axes below. click a point to delete it.

Explanation:

Step1: Recall Exponential Function Transformations

The general form of an exponential function is \( y = a^{x - h} + k \), where the horizontal asymptote is \( y = k \). For \( y = 2^{x - 2} - 5 \), \( k = -5 \), so the horizontal asymptote is \( y = -5 \) (which matches the given boxed value, so we can confirm that part). Now, to find points, we choose integer values for \( x \) and compute \( y \).

Step2: Find Points for \( x = 0 \)

Substitute \( x = 0 \) into the function: \( y = 2^{0 - 2} - 5 = 2^{-2} - 5 = \frac{1}{4} - 5 = -\frac{19}{4} = -4.75 \). Wait, but we need integer coordinates? Wait, maybe I made a mistake. Wait, let's check \( x = 2 \): \( y = 2^{2 - 2} - 5 = 2^{0} - 5 = 1 - 5 = -4 \). So point \( (2, -4) \).

Step3: Find Point for \( x = 3 \)

Substitute \( x = 3 \): \( y = 2^{3 - 2} - 5 = 2^{1} - 5 = 2 - 5 = -3 \). So point \( (3, -3) \).

Step4: Find Point for \( x = 1 \)

Substitute \( x = 1 \): \( y = 2^{1 - 2} - 5 = 2^{-1} - 5 = \frac{1}{2} - 5 = -4.5 \). Not integer. Try \( x = 4 \): \( y = 2^{4 - 2} - 5 = 2^{2} - 5 = 4 - 5 = -1 \). So point \( (4, -1) \).

Step5: Find Point for \( x = -1 \)

Substitute \( x = -1 \): \( y = 2^{-1 - 2} - 5 = 2^{-3} - 5 = \frac{1}{8} - 5 = -4.875 \). Not integer. Wait, \( x = 5 \): \( y = 2^{5 - 2} - 5 = 8 - 5 = 3 \). So point \( (5, 3) \). Wait, maybe the problem allows non-integer? No, the instruction says "integer coordinates". Wait, maybe I misread. Wait, the function is \( y = 2^{x - 2} - 5 \). Let's list more \( x \) values:

• \( x = 2 \): \( y = 1 - 5 = -4 \) (integer)
• \( x = 3 \): \( 2 - 5 = -3 \) (integer)
• \( x = 4 \): \( 4 - 5 = -1 \) (integer)
• \( x = 5 \): \( 8 - 5 = 3 \) (integer)
• \( x = 1 \): \( 2^{-1} - 5 = -4.5 \) (not integer)
• \( x = 0 \): \( 2^{-2} - 5 = -4.75 \) (not integer)
• \( x = -1 \): \( 2^{-3} - 5 = -4.875 \) (not integer)

So the integer-coordinate points are \( (2, -4) \), \( (3, -3) \), \( (4, -1) \), \( (5, 3) \), and also \( x = 2 \) is good, \( x = 2 \), \( x = 3 \), \( x = 4 \), \( x = 5 \), and maybe \( x = -2 \): \( y = 2^{-4} - 5 = 1/16 - 5 = -4.9375 \). No. Wait, maybe the problem has a typo, but the asymptote is \( y = -5 \), which is correct because for exponential functions \( a^{x - h} + k \), the horizontal asymptote is \( y = k \). So the asymptote equation is \( y = -5 \), which is already given as \( y = -5 \). For the points, we can plot \( (2, -4) \), \( (3, -3) \), \( (4, -1) \), \( (5, 3) \), and maybe \( (2, -4) \), \( (3, -3) \), \( (4, -1) \), \( (1, -4.5) \) but no, integer. Wait, maybe the problem allows approximate? No, the instruction says "integer coordinates". So the key points with integer coordinates are \( (2, -4) \), \( (3, -3) \), \( (4, -1) \), \( (5, 3) \), and also \( x = 2 \) is the vertex-like point (where the exponent is 0). So the asymptote is \( y = -5 \), which is correct.

Answer:

The equation of the asymptote is \( y = -5 \), and some integer-coordinate points are \( (2, -4) \), \( (3, -3) \), \( (4, -1) \), \( (5, 3) \) (to plot on the graph). The asymptote equation is \( \boxed{y = -5} \) (already provided, but confirming).