State the domain, range, intercept(s), and asymptote(s) of y equals negative 2 superscript negative x plus 2 baseline plus 1.

(4 points)

1 answer

To analyze the function \( y = -2^{-x+2} + 1 \), we can break it down and identify its domain, range, intercepts, and asymptotes.

Step 1: Identify the Domain

The domain of the function \( y = -2^{-x+2} + 1 \) is all real numbers. Therefore:

  • Domain: \( (-\infty, \infty) \)

Step 2: Identify the Range

The expression \( -2^{-x+2} \) will vary based on the exponent. As \( x \) increases, \( -2^{-x+2} \) approaches \( 0 \) from below since \( 2^{-x+2} \) becomes very small. The function will never actually reach \( 0 \) but can get arbitrarily close to it. The vertical shift upwards by \( 1 \) means that the values of the function approach \( 1 \) from below. However, when \( x \) is very large and \( -x+2 \) becomes negative, \( 2^{-x+2} \) gets larger, and thus \( -2^{-x+2} \) approaches \( -\infty \). Hence:

  • Range: \( (-\infty, 1) \)

Step 3: Identify the Intercepts

Y-Intercept: To find the y-intercept, set \( x = 0 \): \[ y = -2^{-(0)+2} + 1 = -2^{-2} + 1 = -\frac{1}{4} + 1 = \frac{3}{4} \] So, the y-intercept is:

  • Y-Intercept: \( (0, \frac{3}{4}) \)

X-Intercept: To find the x-intercept, set \( y = 0 \): \[ 0 = -2^{-x+2} + 1 \] \[ 2^{-x+2} = 1 \] Taking the log base 2 of both sides: \[ -x + 2 = 0 \implies x = 2 \] Thus, the x-intercept is:

  • X-Intercept: \( (2, 0) \)

Step 4: Identify the Asymptotes

The function has a horizontal asymptote because as \( x \to \infty \), \( y \to 1 \).

  • Horizontal Asymptote: \( y = 1 \)

Summary:

  • Domain: \( (-\infty, \infty) \)
  • Range: \( (-\infty, 1) \)
  • Y-Intercept: \( (0, \frac{3}{4}) \)
  • X-Intercept: \( (2, 0) \)
  • Horizontal Asymptote: \( y = 1 \)