Question

1. $y = 2(x + 3)^2 + 3$

Explanation:

Step1: Identify the vertex form

The equation \( y = 2(x + 3)^2 + 3 \) is in vertex form \( y = a(x - h)^2 + k \), where \((h, k)\) is the vertex.

Step2: Determine h and k

In \( y = 2(x + 3)^2 + 3 \), we can rewrite \( x + 3 \) as \( x - (-3) \), so \( h = -3 \) and \( k = 3 \). Thus, the vertex is \((-3, 3)\).

Step3: Analyze the coefficient a

The coefficient \( a = 2 \), which is positive, so the parabola opens upward.

Step4: Find additional points (optional for graphing)

Let's find some points. For \( x = -2 \):
\( y = 2(-2 + 3)^2 + 3 = 2(1)^2 + 3 = 2 + 3 = 5 \), so the point is \((-2, 5)\).
For \( x = -4 \):
\( y = 2(-4 + 3)^2 + 3 = 2(-1)^2 + 3 = 2 + 3 = 5 \), so the point is \((-4, 5)\).
For \( x = -1 \):
\( y = 2(-1 + 3)^2 + 3 = 2(2)^2 + 3 = 8 + 3 = 11 \), so the point is \((-1, 11)\).
For \( x = -5 \):
\( y = 2(-5 + 3)^2 + 3 = 2(-2)^2 + 3 = 8 + 3 = 11 \), so the point is \((-5, 11)\).

To graph: Plot the vertex \((-3, 3)\), then plot the points \((-2, 5)\), \((-4, 5)\), \((-1, 11)\), \((-5, 11)\), and draw a parabola opening upward through these points.

Answer:

The vertex of the parabola \( y = 2(x + 3)^2 + 3 \) is \((-3, 3)\), it opens upward, and additional points like \((-2, 5)\), \((-4, 5)\), \((-1, 11)\), \((-5, 11)\) can be used to sketch the graph. (If the question was to find the vertex or graph the function, this is the relevant information.)