Question

What are the domain, range, and asymptote of h(x) = (1.4)^x + 5?

Answer 1

Answer:

A.Domain:{x\x is a real number }

Range:{y\y>5}

Asymptote :y=5

Step-by-step explanation:

We are given that a function

$h(x)=(1.4)^x+5$

We have to find the domain, range and asymptote of h(x).

It is exponential function therefore, it is defined for all values of x.

Hence, domain  of h(x)={x| x is a real number}

Substitute x=0 then we get

$h(0)=(1.4)^0+5=1+5=6$

(a^0=1)

Hence, range of h(x)={y|y>5}

For exponential function,

Horizontal asymptote:$a^x\rightarrow 0$ when $x\rightarrow-\infty$

Apply limit $x\rightarrow-\infty$

$\lim_{x\rightarrow-\infty}h(x)=\lim_{x\rightarrow-\infty}(1.4)^x+5=0+5=5$

$e^{-\infty}=0$

$\lim_{x\rightarrow \infty}(1.4)^x+5=\infty$

Hence, the horizontal asymptote y=5.

Answer 2

$(1.4)^x$ is always positive for any real $x$, so by itself the range would be $y>0$, but $(1.4)^x+5$ adds 5 to every number in that original range. This means the actual range would be all positive numbers greater than 5, or $\{y~|~y>5\}$.

Conveniently, only (B) has this as an option for the range, so this must be the answer. (The other two properties check out, since $x$ can indeed be any real number, while $\displaystyle\lim_{x\to\-\infty}h(x)=5$, so $y=5$ is indeed a horizontal asymptote of $h(x)$.)