Question

The linear function p(x) has a slope of negative three-fifths and passes through the ordered pairs (0, 3) and (5, 0). The exponential function q(x) follows a pattern of exponential decay and passes through the ordered pairs (–1, 3) and (0, 1). Which statement accurately describes the end behavior the two graphs have in common?

Answer 1

In the image at the end we can see the graph for both functions, there we can see that:

as x → -∞, y → ∞

Which statement accurately describes the end behavior the two graphs have in common?

Here we have a linear function with a negative slope of 3/5 and we know that it passes through (0, 3), so we can write this as:

y - 3 = (-3/5)*x

And we have an expónential decay which passes through (-1, 3) and (0, 1)

The general exponential is written as:

y = a*bˣ

We know it passes through (0, 1) then:

1 = a*b⁰ = a

y = bˣ

And it passes through (-1, 3), then:

3 = b⁻¹

3 = 1/b

b = 1/3

The exponential is:

y = (1/3)ˣ

In the image below you can see both graphs:

There we can see that the common end behavior for both functions is that:

as x → -∞, y → ∞