Question

match each polynomial function to its graph.
$f(x) = -2x^2 - 6x = -2x(x + 3)$
$g(x) = -x^2 - 1$

Explanation:

Step1: Analyze \( f(x) = -2x^2 - 6x \)

Factor \( f(x) \) as \( -2x(x + 3) \). Set \( f(x) = 0 \):
\( -2x(x + 3) = 0 \implies x = 0 \) or \( x = -3 \).
Thus, \( f(x) \) has x-intercepts at \( x = -3 \) and \( x = 0 \), so it matches the graph with these intercepts (left graph).

Step2: Analyze \( g(x) = -x^2 - 1 \)

Set \( g(x) = 0 \): \( -x^2 - 1 = 0 \implies x^2 = -1 \).
Since \( x^2 = -1 \) has no real solutions, \( g(x) \) has no x-intercepts. The graph with no x-intercepts (right graph, opening downward) matches \( g(x) \).

Answer:

To match the polynomial functions to their graphs:

• For \( f(x) = -2x^2 - 6x = -2x(x + 3) \):
• The roots (x-intercepts) are found by setting \( f(x) = 0 \): \( -2x(x + 3) = 0 \) gives \( x = 0 \) or \( x = -3 \).
• The graph with x-intercepts at \( x = -3 \) and \( x = 0 \) (the left graph) corresponds to \( f(x) \).

• For \( g(x) = -x^2 - 1 \):
• The equation \( -x^2 - 1 = 0 \) simplifies to \( x^2 = -1 \), which has no real roots (so no x-intercepts).
• The graph with no x-intercepts (the right graph, opening downward with vertex below the x-axis) corresponds to \( g(x) \).

So:
\( f(x) = -2x^2 - 6x \) matches the left graph (with x-intercepts at \( -3 \) and \( 0 \)).
\( g(x) = -x^2 - 1 \) matches the right graph (no x-intercepts).