Question

what must be a factor of the polynomial function f(x) graphed on the coordinate plane below?
(graph of a parabola with vertex at (-2, 0), opening downward, intersecting y-axis at (0, -4))
options:
○ x - 4
○ x - 2
○ x + 2
○ x + 4

Explanation:

Step1: Identify the root of the polynomial

The graph of the polynomial touches the x - axis at \(x=- 2\). When a graph touches the x - axis at a point \(x = a\), it means that \((x - a)\) is a factor with an even multiplicity. Here, \(a=-2\), so the factor corresponding to the root \(x=-2\) is found by setting \(x+2 = 0\) (since if \(x=-2\), then \(x + 2=-2 + 2=0\)).

Step2: Determine the factor

For a root \(x = r\) of a polynomial \(f(x)\), the factor is \((x - r)\). In our case, \(r=-2\), so the factor is \((x-(-2))=x + 2\). We can check the other options:

• For \(x-4\), the root would be \(x = 4\), but the graph does not cross or touch the x - axis at \(x = 4\).
• For \(x - 2\), the root would be \(x=2\), but the graph does not cross or touch the x - axis at \(x = 2\).
• For \(x + 4\), the root would be \(x=-4\), but the graph does not cross or touch the x - axis at \(x=-4\) (it touches at \(x=-2\)).

Answer:

\(x + 2\) (corresponding to the option "x + 2")