[tex]x=0,y=7[/tex]
1) In this problem, we can write two t-tables. One t-table and find the roots of the quadratic equation, and the other t-table for the linear equation. We'll then trace the graph and check
2) We can plug random values into the equation so that we can get the values for y:
I) y=x²+2x+7
x|y
-1| y=6
0| 7
1| 10
(-1,6), (0,7), and (1,10)
[tex]\begin{gathered} x^{2}+2x+7 \\ x=\frac{-2\pm\sqrt[]{4-28}}{2}=x_1=-1+\sqrt[]{6}i,x_2=-1-\sqrt[]{6}i \end{gathered}[/tex]
So as we can see, the complex roots tell us that the parabola does not cross the x-axis.
II) y=2x+7
x|y
-1| 5
0| 7
1| 9
(-1, 5), (0,7), and (1,9)
3) Now, we can plot both equations:
Note that the line and the parabola have one common point. So the answer to that Linear System is point (0,7) i.e., x=0 and y=7
