Question

how would you explain the relationship between the real zero(s) of the function and x-intercept(s) of the graph?

since the graph crosses the x-axis at \\(x = -2\\), the function has a real zero of \\(x = -2\\).
since the graph never crosses the x-axis, the function has no real zeros.
since the graph eventually crosses the x-axis, the function has a real zero.
since the graph crosses the y-axis at \\(\frac{1}{9}\\), the function results in a real zero of \\(x = \frac{1}{9}\\)?

Explanation:

Analyze the graph and function

Using the Exponential Function Graphs knowledge point
The graph represents the exponential function \(y = 3^{x-2}\). As \(x \to -\infty\), the graph approaches the horizontal asymptote \(y = 0\) (the \(x\)-axis) but never touches or crosses it.

Determine the existence of x-intercepts

Using the Asymptotic Behavior of Logarithms knowledge point
Because \(3^{x-2} > 0\) for all real numbers \(x\), there is no value of \(x\) that satisfies \(3^{x-2} = 0\). Thus, the graph has no \(x\)-intercepts.

Connect x-intercepts to real zeros

The real zeros of a function correspond to the \(x\)-intercepts of its graph. Since the graph never crosses or touches the \(x\)-axis, the function has no real zeros.

Answer:

• Since the graph crosses the \(x\)-axis at \(x = -2\), the function has a real zero of \(x = -2\).
• Since the graph never crosses the \(x\)-axis, the function has no real zeros. (Correct answer)
• Since the graph eventually crosses the \(x\)-axis, the function has a real zero.
• Since the graph crosses the \(y\)-axis at \(\frac{1}{9}\), the function results in a real zero of \(x = \frac{1}{9}\)?