
Points you need to find to graph quadratic inequalities:
Vertex of each parabola:
1-Write each ineqaulity as an equation:

Vertex:

First equation: the leding coefficient is 1 then the parabola opens up.
Vertex of first equation:

Second equation: the leading coefficient is -1 then the parabola opens down.
Vertex of the second equation:

Points of interception:
Equal the equations and solve x:
![\begin{gathered} x^2-4x+8=-x^2+4x+2 \\ \\ x^2+x^2-4x-4x+8-2=0 \\ 2x^2-8x+6=0 \\ \\ \text{Quadratic formula:} \\ ax^2+bx+c=0 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \\ \\ x=\frac{-(-8)\pm\sqrt[]{(-8)^2-4(2)(6)}}{2(2)} \\ \\ x=\frac{8\pm\sqrt[]{64-48}}{4} \\ \\ x=\frac{8\pm\sqrt[]{16}}{4} \\ \\ x=(8\pm4)/(4) \\ \\ x_1=(8+4)/(4)=(12)/(4)=3 \\ \\ x_2=(8-4)/(4)=(4)/(4)=1 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/3s49hhwkw4u8itjnybgd8dwc0d30kn94zx.png)
The parabolas intersect in x=1 and x=3 (use one of the equations to find the y-value of the intersection):

Then, you have the next points:
Vertex: (2,4) opens up; (2,6) opens down
Intersection points: (1,5) and (3,5)
First parabola has the inequality sing > : the border line is a dotted line and the shadow area is under the parabola.
Second parabola has the inequality sing ≤ : the border line is a full line and the shadow area is over the parabola
Graph: