Question

which graph represents the function $y^2 = -16x$?

Explanation:

Step1: Recall the standard form of a parabola

The standard form of a parabola that opens left or right is \((y - k)^2 = 4p(x - h)\), where \((h,k)\) is the vertex and \(p\) determines the direction and width. For the given equation \(y^2=-16x\), we can rewrite it as \((y - 0)^2=4(-4)(x - 0)\). So the vertex \((h,k)=(0,0)\), and \(p=-4\).

Step2: Determine the direction and vertex

Since \(p=-4<0\), the parabola opens to the left. The vertex is at the origin \((0,0)\). Also, the focus is at \((h + p,k)=(0 - 4,0)=(-4,0)\) and the directrix is \(x=-p = 4\) (wait, no: for \((y - k)^2=4p(x - h)\), if \(p<0\), opens left, directrix is \(x=h - p\)? Wait, no, let's correct: the standard form for a parabola opening left/right is \((y - k)^2=4p(x - h)\). If \(p>0\), opens right; \(p<0\), opens left. The vertex is \((h,k)\), focus is \((h + p,k)\), directrix is \(x=h - p\). So for \(y^2=-16x=(y - 0)^2=4(-4)(x - 0)\), so \(h = 0\), \(k = 0\), \(p=-4\). Then focus is \((0-4,0)=(-4,0)\), directrix is \(x=0-(-4)=4\)? Wait, no, wait: \(4p=-16\), so \(p=-4\). So the formula for directrix when the parabola is \((y - k)^2=4p(x - h)\) is \(x=h - p\)? Wait, no, let's derive it. For a parabola opening to the right, \(y^2 = 4ax\) (where \(a>0\)), vertex at (0,0), focus at (a,0), directrix \(x=-a\). If it's opening to the left, \(y^2=-4ax\) (a>0), then vertex at (0,0), focus at (-a,0), directrix \(x=a\). So in our case, \(y^2=-16x\), so \(4a = 16\), so \(a = 4\). So it's a parabola opening to the left, vertex at (0,0), focus at (-4,0), directrix \(x = 4\). Now let's check the graphs:

- First graph: vertex at (0,0)? Wait, the first graph has a vertex? Wait, the first graph has a parabola opening left, with vertex near the origin? Wait, the first graph has a point (-4,0) which is the focus, and directrix \(x = 4\), and the parabola opens left, passing through (0,4) and (0,-4) (since when \(x = 0\), \(y = 0\); when \(x=-4\), \(y^2=-16*(-4)=64\), so \(y=\pm8\)? Wait, no, wait when \(x=-4\), \(y^2=-16*(-4)=64\), so \(y=\pm8\)? Wait, maybe my earlier thought was wrong. Wait, let's take the equation \(y^2=-16x\). Let's find some points. When \(x = 0\), \(y = 0\). When \(x=-1\), \(y^2=16\), so \(y=\pm4\). When \(x=-4\), \(y^2=64\), so \(y=\pm8\). So the parabola passes through (0,0), (-1,4), (-1,-4), (-4,8), (-4,-8). Now let's check the graphs:

First graph: has a parabola opening left, with vertex at (0,0)? Wait, the first graph has a point (-4,0) (the focus), directrix \(x = 4\), and the parabola passes through (0,4) and (0,-4)? Wait, no, when \(x = 0\), \(y = 0\), but in the first graph, when \(x = 0\), \(y\) can be \(\pm4\)? Wait, maybe the first graph is the correct one. Let's check the other graphs:

Second graph: has vertex at (-16,0), which is wrong, since our vertex is at (0,0). Third graph: opens up/down? No, it's a horizontal parabola? Wait, no, the third graph has horizontal lines \(y = -16\) and \(y = 16\), so it's a vertical parabola? No, the third graph's parabola is opening up/down? Wait, no, the equation \(y^2=-16x\) is a horizontal parabola (opens left/right), so the third and fourth graphs are vertical parabolas (open up/down), so they can be eliminated. Now between first and second graph: second graph has vertex at (-16,0), which is wrong. First graph: vertex at (0,0) (since when \(x = 0\), \(y = 0\)), focus at (-4,0) (the purple dot), directrix \(x = 4\) (the dashed line), and the parabola opens left, passing through (0,4) and (0,-4)? Wait, when \(x = 0\), \(y = 0\), but when \(x=-4\), \(y^2=64\), so \(y=\pm8\), but maybe the graph is scaled. Wait, the first graph has a grid where each square is 1 unit? Wait, the first graph: x-axis from -4 to 4, y-axis from -4 to 4. So when \(x=-1\), \(y^2=16\), so \(y=\pm4\), which would be on the graph (since at \(x=-1\), y=4 and y=-4, which are on the first graph's parabola). So the first graph matches: opens left, vertex at (0,0), focus at (-4,0), directrix \(x = 4\), and passes through (0,4) and (0,-4) (when \(x=-1\), but maybe the graph is drawn with \(x=-4\) as the focus, etc. So the first graph is the correct one.

Answer:

The first graph (the one with vertex near the origin, focus at (-4,0), directrix \(x = 4\), and the parabola opening left).