Answer
See the graph
Work Step by Step
Given: $y=2x^2+6x+3$
The coefficients are $a =2$, $b =6$, and $c=3$. Because $a > 0$, the parabola opens up.
Find the vertex.
$x=-\frac{b}{2a}=\frac{-6}{2.2}=-\frac{3}{2}$
Then find the y-coordinate of the vertex.
$y=2(-\frac{3}{2})^2+6(-\frac{3}{2})+3=-\frac{3}{2}$
Draw the axis of symmetry $x =-\frac{3}{2}$
The y-intercept is $3$. Plot the point $(0, 3)$. Then reflect this point in the axis of symmetry to plot another point, $(-3,3)$.
Evaluate the function for another value of $x$, such as $x =-4$.
$y=2(-4)^2+6(-4)+3=11$
Plot the point $(-4, 11)$ and its reflection $(1, 11)$ in the axis of symmetry.
Draw a parabola through the plotted points.
