[tex]y=x^2+2x+7[/tex]
Vertex:
[tex]\begin{gathered} f(x)=ax^2+bx+c \\ \\ (-\frac{b}{2a},f(-\frac{b}{2a})) \end{gathered}[/tex]
For the given function:
[tex]\begin{gathered} x-coordinate \\ -\frac{b}{2a}=-\frac{2}{2(1)}=-1 \\ \\ y-coordinate \\ f(-1)=(-1)^2+2(-1)+7 \\ f(-1)=1-2+7 \\ f(-1)=6 \end{gathered}[/tex]
Then, the vertex of the given parabola is the point (-1,6)
As the leading coefficient of the parabola (a) is greather than 0 the parabola opens up. When the parabola opens up and the y-coordinate of the vertex is greather than 0 the function has not real zeros.
As the vertex is the minimum value in a parabola that opens up you need to find points that have greather values of y than the vertex. Or you can find one point on the left (x-coordinate less than -1) of the vertex and one on the right (x-coordinate greather than -1)
Point A:
When x= -3
[tex]\begin{gathered} y=(-3)^2+2(-3)+7 \\ y=9-6+7 \\ y=10 \end{gathered}[/tex]
Point A (-3, 10)
Point B:
when x=0
[tex]\begin{gathered} y=0^2+2(0)+7 \\ y=0+0+7 \\ y=7 \end{gathered}[/tex]
Point B: (0,7)
You get the next graph.
