Question

question 25 of 25
which direction does the graph of the equation shown below open?
$y^{2}+16y - 4x + 4 = 0$
a. up
b. right
c. down
d. left

Explanation:

Step1: Rewrite the equation in standard form

The standard form of a parabola is \((y - k)^2 = 4p(x - h)\) (opens left or right) or \((x - h)^2 = 4p(y - k)\) (opens up or down).
For the given equation \(y^{2}+16y - 4x + 4 = 0\), complete the square for the \(y\) - terms.

$$
LATEXBLOCK0
$$

Step2: Determine the direction of the parabola

In the standard form \((y - k)^2 = 4p(x - h)\), if \(p>0\), the parabola opens to the right; if \(p < 0\), it opens to the left.
Comparing \((y + 8)^{2}=4(x + 15)\) with \((y - k)^2 = 4p(x - h)\), we have \(4p = 4\), so \(p = 1>0\).

So the parabola opens to the right.

Answer:

B. Right