Question

find the vertex, focus, and directrix of the parabola without completing the square, and determine whether the parabola opens upward or downward.
( y=\frac{1}{16} x^{2}-5 )
the coordinates of the vertex are ( (0,-5) ).
(simplify your answer. type an ordered pair.)
the coordinates of the focus are ( (0,-1) ).
(simplify your answer. type an ordered pair.)
the directrix is (
abla=square ).
(simplify your answer.)

Explanation:

Step1: Recall the standard form of parabola

The standard form of a parabola that opens up or down is \(x^{2}=4p(y - k)\), which can be rewritten as \(y=\frac{1}{4p}(x - h)^{2}+k\). For the given equation \(y=\frac{1}{16}x^{2}-5\), we have \(h = 0\), \(k=-5\), and \(\frac{1}{4p}=\frac{1}{16}\).

Step2: Solve for \(p\)

From \(\frac{1}{4p}=\frac{1}{16}\), cross - multiply gives \(4p = 16\), so \(p = 4\).

Step3: Find the directrix

The formula for the directrix of a parabola \(y=\frac{1}{4p}(x - h)^{2}+k\) is \(y=k - p\). Substitute \(k=-5\) and \(p = 4\) into the formula. Then \(y=-5-4=-9\).

Answer:

The directrix is \(y=-9\)