As we can see, x has the be less than 13. In a diagram, we have to mark the endpoint as an open circle as it is a strict inequality. Since the inequality represents all x less than 13, we should now shade everything to the left of 13.
b
To determine the solution set we will first have to find its boundary point(s) by treating the inequality as an equation.
Inequality:& 3x^2-4x+2≤ x^2+x+6
Equation:& 3x^2-4x+2= x^2+x+6
If we solve this equation, we can determine the boundary points. To solve it, we will start by setting one side equal to 0.
The boundary points are located at x=-0.64 and x=3.14. Let's mark them on a number line. Since the inequality is non-strict, the boundaries are not included in the solution set. To determine where we should shade the number line, we will also include three test points.
Now, substitute x in the inequality with these values and see which one holds.
|c|c|c|
[-0.8em]
x & 3x^2-4x+2≤ x^2+x+6 & Evaluate [0.5em]
[-1em]
-1 & 3( -1)^2-4( -1)+2? ≤ ( -1)^2+( -1)+6 & 9 ≰ 6 * [0.5em]
[-1em]
0 & 3( 0)^2-4( 0)+2? ≤ 0^2+ 0+6 & 2 ≤ 6 ✓ [0.5em]
[-1em]
4 & 3( 4)^2-4( 4)+2? ≤ 4^2+ 4+6 & 34 ≰ 26 * [0.5em]
The inequality is true to the left and right of the boundary points. Therefore, we should shade this region.
