Question

question 6 of 10
which describes the graph of $y = (x + 3)^2 - 4$?

a. opens down with a vertex at $(-3, -4)$

b. opens up with a vertex at $(3, -4)$

c. opens down with a vertex at $(3, -4)$

d. opens up with a vertex at $(-3, -4)$

Explanation:

Step1: Recall the vertex form of a parabola

The vertex form of a parabola is \( y = a(x - h)^2 + k \), where \((h, k)\) is the vertex and \(a\) determines the direction it opens. If \(a>0\), it opens up; if \(a<0\), it opens down.

Step2: Identify \(a\), \(h\), and \(k\) from the given equation

For the equation \( y=(x + 3)^2-4 \), we can rewrite \(x + 3\) as \(x-(-3)\). So, comparing with \( y = a(x - h)^2 + k \), we have \(a = 1\) (since there's no coefficient other than 1 in front of \((x + 3)^2\)), \(h=-3\), and \(k = - 4\).

Step3: Determine the direction and vertex

Since \(a = 1>0\), the parabola opens up. The vertex is \((h,k)=(-3,-4)\).

Answer:

D. Opens up with a vertex at \((-3, -4)\)