Question

Graphing Exponential Function in Exercise ,sketch the graph of the function.See example 3 and 4.

Answer 1

Answer:

The graph is shown below.

Step-by-step explanation:

Given:

The function to graph is given as:

$y=2^{-x^{2}}$

Let us first find the key points of the above equation.

1. Horizontal asymptotes:

Horizontal asymptotes are the values of 'y' for which 'x' tends to infinity.

So, we obtain the horizontal asymptotes from the limit.

$\lim_{x \to \infty} 2^{-x^2}= \lim_{x \to \infty} \frac{1}{2^{x^2}}=\frac{1}{\infty}=0\\\\ \lim_{x \to- \infty} 2^{-x^2}= \lim_{x \to- \infty} \frac{1}{2^{x^2}}=\frac{1}{-\infty}=0$

So, the horizontal asymptotes occur at the line $y=0$ or x-axis.

2. The given function is an even function as there is square of 'x'.

So, an even function is a function that is symmetric about y-axis.

Therefore, the graph of the given function is symmetric about y-axis.

3. y-intercept:

The point on the y-axis where the graph crosses it is the y-intercept.

Plug in 0 for 'x' and find 'y'. This gives,

$y=2^{0}=1$. So, the point is (0, 1)

4. Now, finding two more points on the graph in order to graph it.

Let $x=-1\ and\ x=1$. Find 'y'.

$y=2^{-(-1)^2}=2^{-1}=\frac{1}{2}=0.5\\\\y=2^{-(1)^2}=2^{-1}=\frac{1}{2}=0.5$

Therefore, two more points on the graph are (-1, 0.5) and (1, 0.5)

Now, let us plot the graph using the above information.

Plot the left half of the graph and then reflect the same on the right half.

The graph is shown below.