Graph y< 1/3 c + 1/2
1 answer:
<h3>The Graph you need above </h3><h3>Step-by-step explaination : </h3><h3>Combine 1/3and x.</h3>
y<x/3+1/2
<h3>Use the slope-intercept form to find the slope and y-intercept.</h3><h3 />Slope: 1/3
y-intercept: 1/2
<h3>Graph a dashed line, then shade the area below the boundary line since y is less</h3><h3>than x/3+1/2.</h3><h3>y<x/3+1/2</h3>
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Check the picture below.
now, we know the directrix is at y = 1, and the focus point is at 1,3, well, notice the picture, the distance between those fellows is just 2 units.
the vertex is half-way between those fellows, therefore, the vertex will be at 1,2.
the distance "p", from the vertex to either the directrix or focus, is really just 1 unit. Since the focus point is above the directrix, is a vertical parabola, and it opens upwards, like in the picture, and since it opens up, the "p" value is positive, or +1.
now, we know the directrix is at y = 1, and the focus point is at 1,3, well, notice the picture, the distance between those fellows is just 2 units.
the vertex is half-way between those fellows, therefore, the vertex will be at 1,2.
the distance "p", from the vertex to either the directrix or focus, is really just 1 unit. Since the focus point is above the directrix, is a vertical parabola, and it opens upwards, like in the picture, and since it opens up, the "p" value is positive, or +1.