Find the slope of a line perpendicular to y = 5/2x +1 A.-2/5 B.2/5 C.-5/2 D. 5/2
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So hmm notice the graph below
based on where the focus point is at, and the directrix, then, the parabola is opening upwards, meaning the squared variable is the "x"
now, keep in mind, the vertex is at coordinates h,k
the vertex itself is half-way between the focus and directrix
the directrix is at y=15/8 and the focus is at y=17/8
so, half-way will then be
well, so is 1/4 between the focus point and the directrix, half of that is 1/8
so, if you move from the focus point 1/8 down, you'll get the y-coordinate for the vertex, or 1/8 up from the directrix, since the vertex is equidistant to either
what's the "p" distance? well, we just found it, is just 1/8
so, the x-coordinate is obviously -4, get the y-coordinate by 17/8 - 1/8 or 15/8 + 1/8
and plug your values (x-h)² = 4p(y-k) and then solve for "y", that's the equation of the quadratic
based on where the focus point is at, and the directrix, then, the parabola is opening upwards, meaning the squared variable is the "x"
now, keep in mind, the vertex is at coordinates h,k
the vertex itself is half-way between the focus and directrix
the directrix is at y=15/8 and the focus is at y=17/8
so, half-way will then be
well, so is 1/4 between the focus point and the directrix, half of that is 1/8
so, if you move from the focus point 1/8 down, you'll get the y-coordinate for the vertex, or 1/8 up from the directrix, since the vertex is equidistant to either
what's the "p" distance? well, we just found it, is just 1/8
so, the x-coordinate is obviously -4, get the y-coordinate by 17/8 - 1/8 or 15/8 + 1/8
and plug your values (x-h)² = 4p(y-k) and then solve for "y", that's the equation of the quadratic