To plot the equation $y = x^2 + 2x + 1$, we will create a table of values and then draw the corresponding points on a graph.

First, let’s analyze the equation $y = x^2 + 2x + 1$. This is a quadratic equation, which means its graph will be a parabola. The general form of a quadratic equation is given by:

where in this case, we have $a = 1$, $b = 2$, and $c = 1$.

Next, we will create a table of values to assist in plotting the graph. We will select a range of $x$ values that include both negative and positive numbers and calculate the corresponding $y$ values. Here’s the table:

\begin{array}{c|c} x & y = x^2 + 2x + 1 \ \hline -3 & (-3)^2 + 2(-3) + 1 = 9 - 6 + 1 = 4 \ -2 & (-2)^2 + 2(-2) + 1 = 4 - 4 + 1 = 1 \ -1 & (-1)^2 + 2(-1) + 1 = 1 - 2 + 1 = 0 \ 0 & (0)^2 + 2(0) + 1 = 0 + 0 + 1 = 1 \ 1 & (1)^2 + 2(1) + 1 = 1 + 2 + 1 = 4 \ 2 & (2)^2 + 2(2) + 1 = 4 + 4 + 1 = 9 \ \end{array}

Now, we can plot these points on a graph with an $x$-axis and a $y$-axis. The points to plot are as follows: $(-3, 4)$, $(-2, 1)$, $(-1, 0)$, $(0, 1)$, $(1, 4)$, and $(2, 9)$.

Once you have plotted these points, draw a smooth curve connecting them. You will observe that the graph forms a U-shaped parabola that opens upwards. The vertex of this parabola, which represents its lowest point, is located at the coordinates $(-1, 0)$. This can be confirmed by rewriting the equation in vertex form as follows:

This shows that the vertex occurs at $x = -1$, confirming that the lowest point of the parabola is indeed at $(-1, 0)$.