Question

Which is the graph of g(x) = (0.5)x + 3 – 4?

On a coordinate plane, an exponential function decreases in quadrant 2 and has a horizontal asymptote at y = negative 4. It crosses the y-axis at (0, negative 4).
On a coordinate plane, an exponential function decreases in quadrant 2 and has a horizontal asymptote at y = 3. It crosses the y-axis at (0, 3).
On a coordinate plane, an exponential function decreases in quadrant 2 and has a horizontal asymptote at y = negative 4. It crosses the y-axis at (0, 4).
On a coordinate plane, an exponential function decreases in quadrant 2 and has a horizontal asymptote at y = 4. It crosses the y-axis at (0, 12).

Answer 1

I need one clarification — is the function exponential (with an exponent) or linear? The way you wrote it, g(x) = (0.5)x + 3 – 4, reads like a linear function g(x) = 0.5x − 1. If it's exponential you probably meant one of these common forms:
- g(x) = (1/2)^x + 3 − 4 = (1/2)^x − 1, or
- g(x) = (1/2)^{x+3} − 4, or
- something like a·(1/2)^x + k.

A quick test that eliminates two choices right away: an exponential of the form a·(1/2)^x + k has horizontal asymptote y = k and y-intercept g(0) = a + k. So if the asymptote and the y-intercept are the same (choices 1 and 2), that would force a = 0 and the function would be constant — not a decreasing exponential. So choices 1 and 2 are impossible.

Which of the remaining two is correct depends on the exact g(x):
- Option 3 (asymptote y = −4, crosses y-axis at (0,4)) matches g(x) = 8·(1/2)^x − 4 (here k = −4 and a = 8 so g(0)=8−4=4).
- Option 4 (asymptote y = 4, crosses y-axis at (0,12)) matches g(x) = 8·(1/2)^x + 4 (k = 4, a = 8 so g(0)=8+4=12).

Tell me the exact intended formula (was there a caret/exponent? was it x+3 in the exponent?), and I’ll pick the correct option.