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