Question

how many real solutions does $g(x)=-1.6 - 0.2x^{2}$ have, based on its graph? (1 point) the equation $g(x)=-1.6 - 0.2x^{2}$ has no real solutions because it has no y - intercepts. the equation $g(x)=-1.6 - 0.2x^{2}$ has one real solution because it crosses the y - axis once. the equation $g(x)=-1.6 - 0.2x^{2}$ has no real solutions because it has no x - intercepts. the equation $g(x)=-1.6 - 0.2x^{2}$ has no real solutions because it crosses the x - axis once.

Explanation:

Step1: Recall real - solution concept

The real solutions of a function \(y = g(x)\) are the \(x\) - values where \(g(x)=0\), which are the \(x\) - intercepts of the graph of the function.

Step2: Analyze the graph

The graph of \(g(x)=-1.6 - 0.2x^{2}\) does not cross the \(x\) - axis. When \(g(x) = 0\), we have the equation \(-1.6-0.2x^{2}=0\), or \(0.2x^{2}=- 1.6\), \(x^{2}=-8\), which has no real solutions. Since the graph has no \(x\) - intercepts, the function \(g(x)\) has no real solutions.

Answer:

The equation \(g(x)=-1.6 - 0.2x^{2}\) has no real solutions because it has no \(x\) - intercepts.