Question

question 4 of 25
which direction does the graph of the equation shown below open?
y² + 16y - 4x + 4 = 0
a. down
b. right
c. up
d. left

Explanation:

To determine the direction the graph of the equation \( y^2 + 16y - 4x + 4 = 0 \) opens, we first rewrite it in the standard form of a parabola.

Step 1: Complete the square for the \( y \)-terms

Start with the given equation:
\[
y^2 + 16y - 4x + 4 = 0
\]
Isolate the \( y \)-terms:
\[
y^2 + 16y = 4x - 4
\]
Complete the square for \( y^2 + 16y \). Take half of 16 (which is 8), square it (which is 64), and add it to both sides:
\[
y^2 + 16y + 64 = 4x - 4 + 64
\]
Factor the left side as a perfect square and simplify the right side:
\[
(y + 8)^2 = 4x + 60
\]
We can also factor out a 4 from the right side:
\[
(y + 8)^2 = 4(x + 15)
\]

Step 2: Identify the type and direction of the parabola

The standard form of a parabola that opens horizontally is:
\[
(y - k)^2 = 4p(x - h)
\]

• If \( p > 0 \), it opens to the right.
• If \( p < 0 \), it opens to the left.

In our equation \( (y + 8)^2 = 4(x + 15) \), we can compare it to the standard form:

• \( h = -15 \)
• \( k = -8 \)
• \( 4p = 4 \), so \( p = 1 \)

Since \( p = 1 > 0 \), the parabola opens to the right.

Answer:

B. Right