Question

question 11
match the function with its graph.
$f(x) = -2(x + 4)^2$

Explanation:

Step1: Identify the vertex form of a parabola

The function is given in vertex form \( f(x) = a(x - h)^2 + k \), where \((h, k)\) is the vertex and \(a\) determines the direction and width. For \( f(x) = -2(x + 4)^2 \), we can rewrite it as \( f(x) = -2(x - (-4))^2 + 0 \), so the vertex is \((-4, 0)\) and \(a = -2\) (negative, so the parabola opens downward).

Step2: Analyze the vertex and direction

• The vertex is at \(x = -4\), so we look for the graph with vertex at \(x = -4\) (left side of the y - axis).
• Since \(a=-2<0\), the parabola opens downward.

Looking at the graphs:

• The first graph (top - left) has a vertex at \(x=-4\) and opens downward. The second graph has a vertex at \(x = - 4\) but let's check the \(a\) value. The coefficient \(|a| = 2\), which means it should be narrower or wider? Wait, the first graph with vertex at \(x=-4\) and opening downward: let's check the value of \(a\). For \(y=-2(x + 4)^2\), when \(x=-3\), \(y=-2(-3 + 4)^2=-2(1)^2=-2\). Let's check the first graph: when \(x=-3\), the y - value should be - 2. The first graph (top - left) at \(x=-3\) (one unit to the right of the vertex \(x=-4\)) has a y - value of - 2 (since from the vertex \((-4,0)\), moving right 1 unit, \(y=-2(1)^2=-2\)). The second graph at \(x=-3\) would have a different y - value if \(a\) was different. Also, the third and fourth graphs have vertices at \(x = 4\) (right side of the y - axis), so they can be eliminated. The first graph (top - left) has the correct vertex \((-4,0)\) and opens downward (consistent with \(a=-2<0\)).

Answer:

The graph with vertex at \((-4,0)\) and opening downward (the first graph in the top - left corner of the given set of graphs).