Answer
The required graph is as follows:
Work Step by Step
The given equation $y=x^2+4x+6$ is a parabola $(\cup-\text{shaped})$ curve.
For y-intercept, $x=0$, so, we have $y=6$. Thus, $(0, 6)$ is y-intercept.
The same height appears twice due to vertical symmetry, so, $y=6$ occurs for any other point as well. Now, $6=x^2+4x+6$, or, $x^2+4x=0$, or, $x(x+4)=0$, which gives $x=0$ and $x=-4$. This gives another point the parabola, i.e., $(-4, 6)$.
The line of symmetry passes from mid, so, $x=-2$ is the line of symmetry.
The minimum point lies on line of symmetry, so, $y=(-2)^2+4(-2)+6=4-8+6=2$, i.e., $(-2, 2)$ is the minimum point.
Now, draw $\cup-\text{shaped}$ curve passing through $(0, 6)$, $(-4, 6)$ and $(-2,2)$ to obtain the following graph.
