Prove that x²+6x+18 is always greater thanfor any value of x.

One way to solve this would be to graph both y=x^2+6x+18 and y=2 - 1/x.  If the graph of the polynomial is always higher up (above) the graph of y=2-1/x, that alone is sufficient cause to state that the poly is always greater than 2-1/x.

You could also do this algebraically:  write the inequality

x^2+6x+18 > 2 -1/x.  This can be rewritten as x^2+6x+18-2+1/x > 0, or
                                                                         x^2+6x+16+1/x > 0

Try x=-3.  Then 9-18+16-1/3 = 6 2/3, which is greater than 0.

Or you could graph x^2+6x+16+1/x by hand or on a calculator.  Is the graph always above the x-axis?  If so,  x²+6x+18 is always greater than 2-1/x.

For me i think it'll land on blue

Do you need with the x value for the minimum? its always between the two intercepts for a parabola