<p>Explain why: No matter what the value of x, x²-8x+22 is never less then 6.</p>
<p>The x is squared and the y-intercept is 22. If you still don’t get it, just plug in values and you’ll see that the x^2 is dominant. Therefore the only place where it could be less than six is at very small number, but since the y-intercept is 22, any small, negative or positive number is greatly increased.</p>
<p>Also do your homework by yourself in the future, or at least say what you don’t understand.</p>
<p>just wanted to know how to explain it thats all but anyways thanx</p>
<p>if youre in calc u can use the derivative test.
if f(x)=x^2-8x+22, then f '(x)=2x-8.
when u set the derivative equal to zero (0=2x-8), x=4.
therefore, the minimum value for the equation is at x=4.
so, plug in 4 into the original equation and you get f(4)=6, which is why 6 is the minimum value.
VIOLA! (takes a bow)</p>