The graph of $ y \equal{} 2x^2 \plus{} 4x \plus{} 3$ has its:

$ \textbf{(A)}\  \text{lowest point at } {(\minus{}1,9)}\qquad 
\textbf{(B)}\ \text{lowest point at } {(1,1)}\qquad \\
\textbf{(C)}\ \text{lowest point at } {(\minus{}1,1)}\qquad 
\textbf{(D)}\ \text{highest point at } {(\minus{}1,9)}\qquad \\
\textbf{(E)}\ \text{highest point at } {(\minus{}1,1)}$
1. The given function is \( y = 2x^2 + 4x + 3 \). This is a quadratic function of the form \( y = ax^2 + bx + c \), where \( a = 2 \), \( b = 4 \), and \( c = 3 \).

2. Since \( a = 2 > 0 \), the parabola opens upwards, indicating that it has a minimum point (vertex).

3. The x-coordinate of the vertex of a parabola given by \( y = ax^2 + bx + c \) can be found using the formula:
   \[
   x = -\frac{b}{2a}
   \]
   Substituting \( a = 2 \) and \( b = 4 \):
   \[
   x = -\frac{4}{2 \cdot 2} = -\frac{4}{4} = -1
   \]

4. To find the y-coordinate of the vertex, substitute \( x = -1 \) back into the original equation:
   \[
   y = 2(-1)^2 + 4(-1) + 3
   \]
   Simplifying this:
   \[
   y = 2(1) + 4(-1) + 3 = 2 - 4 + 3 = 1
   \]

5. Therefore, the vertex of the parabola, which is the lowest point, is at \((-1, 1)\).

6. From the given options, the correct answer is:
   \[
   \boxed{(\text{C}) \ \text{lowest point at } (\minus{}1,1)}
   \]