How do you graph, find any intercepts, domain and range of f(x)=4(1/2)^(x-1)-3?

Feb 6, 2018

Explanation:

Let's think...

To find the domain, we ask ourselves...
What values of $x$ will make the function invalid?

We see that any real number will make the function be valid.

Therefore, the domain is all real numbers.

What is the largest and the smallest possible values you could obtain as the $y$ value?

Well, as $x$ grows larger and larger, we are making $4 {\left(\frac{1}{2}\right)}^{x - 1}$ reach closer and closer to 0.

Essentially, we are heading towards -3 without ever reaching it.

Now, as $x$ gets smaller and smaller, we see that the $y$ values increases without particular limit.

Therefore, the range is $\left(- 3 , \infty\right)$

To find the intercepts, we note the following:

On the y-intercept, the x value is 0.

On the x-intercept, the y value is 0.

Therefore, on the y-intercept, we have:

$y = 4 {\left(\frac{1}{2}\right)}^{0 - 1} - 3$

=>$y = 4 {\left(\frac{1}{2}\right)}^{- 1} - 3$

=>$y = 5$

The y-intercept is at $\left(0 , 5\right)$

Similarly, on the x-intercept, we have:

$0 = 4 {\left(\frac{1}{2}\right)}^{x - 1} - 3$

=>$3 = 4 {\left(\frac{1}{2}\right)}^{x - 1}$

=>$\frac{3}{4} = {\left(\frac{1}{2}\right)}^{x} / \left(\frac{1}{2}\right)$

=>$\frac{3}{4} = 2 {\left(\frac{1}{2}\right)}^{x}$

=>$\frac{3}{8} = {\left(\frac{1}{2}\right)}^{x}$

=>${\log}_{\frac{1}{2}} \left(\frac{3}{8}\right) = x$

=>$1.415 \approx x$

The x-intercept is at $\left(1.415 , 0\right)$

As to graphing it, I recommend plugging in $x$ values of -5,-4,-3,-2,-1,0,1,2,3,4,5 into the function, get the $y$ values, plot the points, and connect them.