The given function is:

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

This equation is already in vertex form, which is generally given by:

$y = a(x - h)^2 + k$

where $(h, k)$ represents the vertex of the parabola, and $a$ determines the direction and width of the parabola.

In this case:

• $a = 1$
• $h = -2$
• $k = 1$

Vertex

The vertex of the parabola is $(-2, 1)$.

Additional Properties

• Since $a = 1$, the parabola opens upwards.
• The vertex is the minimum point of the parabola since it opens upwards.

So, in vertex form, the function confirms that the vertex is $(-2, 1)$.

Would you like further details, or do you have any questions?

Here are some additional questions for practice:

1. What is the axis of symmetry for the function $y = (x+2)^2 + 1$?
2. How does the value of $a$ affect the width of the parabola?
3. How would the graph change if $a$ were -1 instead of 1?
4. What happens to the vertex if we change $(x + 2)^2$ to $(x - 3)^2$?
5. How would you rewrite the equation if the vertex were at $(3, -2)$?

Tip: The vertex form is particularly useful for identifying the maximum or minimum points of a parabola easily.