Question

identify the graph of $q(x) = -(x + 4)^2 + 7$.

Explanation:

Step1: Analyze the parabola's direction

The function is \( q(x) = -(x + 4)^2 + 7 \). The coefficient of the squared term is \(-1\), which is negative. So, the parabola opens downward. This eliminates the first graph (which opens upward).

Step2: Find the vertex

The vertex form of a parabola is \( y = a(x - h)^2 + k \), where \((h, k)\) is the vertex. For \( q(x) = -(x + 4)^2 + 7 \), we can rewrite \( x + 4 \) as \( x - (-4) \), so \( h = -4 \) and \( k = 7 \). The vertex is \((-4, 7)\). Now, check the remaining graphs (the ones opening downward). The third graph has a vertex around \( x = 4 \), \( y = 8 \), which doesn't match. The fourth graph: let's check the vertex's \( x \)-coordinate. The vertex should be at \( x = -4 \). Looking at the fourth graph, the vertex is around \( x = -4 \) (midway between -6 and -2, which is -4) and \( y \)-coordinate around 7? Wait, let's re - check the second graph. Wait, maybe I mislabeled. Wait, the second graph (top right) has a vertex? Wait, no, let's re - evaluate. Wait, the function is \( q(x)=-(x + 4)^2+7\). Let's find the \( y \)-intercept. When \( x = 0 \), \( q(0)=-(0 + 4)^2+7=-16 + 7=-9\). So the \( y \)-intercept is \((0,-9)\). Now, check the graphs:

• First graph: opens upward, \( y \)-intercept positive, eliminate.
• Second graph (top right): \( y \)-intercept at \((0,-20)\)? No, wait, when \( x = 0 \), our function is -9. Wait, maybe the fourth graph (bottom right): when \( x = 0 \), \( y=-9\)? Let's check the vertex \( x=-4\). The bottom right graph has vertex at \( x=-4 \) (since it's symmetric around \( x=-4 \), between -6 and -2), and \( y \)-intercept at \( x = 0 \), \( y\) around -9? Wait, maybe the correct graph is the bottom right one? Wait, no, let's re - do the vertex and direction. The parabola opens downward (because \( a=-1<0 \)), vertex at \((-4,7)\). Let's check the \( x \)-intercepts. Set \( q(x)=0\):

\(-(x + 4)^2+7 = 0\)

\((x + 4)^2=7\)

\(x+4=\pm\sqrt{7}\)

\(x=-4\pm\sqrt{7}\approx-4\pm2.645\), so \( x\approx-4 + 2.645=-1.355\) and \( x\approx-4-2.645=-6.645\). So the \( x \)-intercepts are around \( x=-6.645\) and \( x=-1.355\). Now, looking at the bottom right graph: it has \( x \)-intercepts around -6 and -2 (wait, -6.645 is close to -6, -1.355 is close to -1.3, which is between -2 and 0). And the \( y \)-intercept at \( x = 0 \) is -9. So the bottom right graph (the fourth one) matches the \( x \)-intercepts (around -6.6 and -1.4), the \( y \)-intercept at -9, and the vertex at \((-4,7)\) (since it's a downward - opening parabola with vertex at \( x=-4 \)). The second graph (top right) has a \( y \)-intercept too low (-20), the third graph (bottom left) has vertex at \( x = 4 \), which is wrong. So the correct graph is the bottom right one (the fourth graph, the one with \( x \)-axis from -8 to 0, \( y \)-axis from -12 to 12, vertex at \( x=-4 \), \( y\) around 7, \( y \)-intercept at \((0,-9)\) and \( x \)-intercepts around -6.6 and -1.4).

Wait, maybe I made a mistake in the initial graph labeling. But the key steps are:

1. Direction: opens downward (so eliminate upward - opening graphs).
2. Vertex at \((-4,7)\).
3. \( y \)-intercept at \((0,-9)\).
4. \( x \)-intercepts at \( x=-4\pm\sqrt{7}\approx-6.645\) and \( x\approx-1.355\).

Among the given graphs, the bottom right graph (the fourth one, the one with \( x \)-axis from -8 to 0, \( y \)-axis from -12 to 12) matches these characteristics.

Answer:

The bottom right graph (the fourth graph in the given set, with \( x \)-axis ranging from -8 to 0 and \( y \)-axis from -12 to 12, having a vertex at \( x = - 4 \), \( y \)-intercept at \( (0,-9) \) and \( x \)-intercepts around \( x=-6.6 \) and \( x=-1.4 \)).