Question

Sketch the graph of the function. $$ y=2^{-x^{2}} $$

Short answer

The graph of \(y=2^{-x^{2}}\) is a downward-opening parabola, symmetrical around the \(y\)-axis, with vertex at \((0,1)\), and the x-axis as its horizontal asymptote.

Step-by-step solution

Identify the Type of Function

The given function is an exponential function. Exponential functions have a general form of \(y = a \cdot b^{x}\), where \(a\) and \(b\) are constants. In our case, \(a = 1\) and \(b = 2^{-x^{2}}\). The exponent is a quadratic function, which will affect the graph's shape.

Find the Symmetry

Determine if the function is even, odd, or neither. Since \(f(-x) = f(x)\), the function is even and symmetrical about the \(y\)-axis.

Calculate the Limits

Calculate the limit of the function as \(x\) approaches \(\pm \infty\). Both \(\lim_{{x \to \infty}} 2^{-x^{2}}\) and \(\lim_{{x \to -\infty}} 2^{-x^{2}}\) are zero. This suggests that the \(x\)-axis (i.e, \(y = 0\)) is a horizontal asymptote.

Find the Vertex

The vertex of the function will be when \(x = 0\). If you substitute \(x = 0\) into the function, you'll get \(f(0) = 2^{0}\) or \(f(0) = 1\). So the vertex of the function is at \((0, 1)\).

Sketch the Graph

Combine the information from above steps to draw the graph. Start by plotting the vertex at \((0, 1)\). Then, remember that the graph is symmetric about the \(y\)-axis. As \(x\) moves away from 0 in either direction, the function will approach the \(x\)-axis but never touch it because of the horizontal asymptote. So, draw the graph smoothly going down and approaching the \(x\)-axis as \(x\) gets large and small.