Here's the question you clicked on:
talleymarks
Determine where f(x) intersects the x-axis. f(x)=(x+10)2−81
the 2 is supposed to me an exponant...
intersecting x-axis implies y-coordinate is zero, therefore eqating f(x)=0 ew get \[0=(x+2)^{2}-81\] \[x+2=\pm9\] x=-11,7 So the given fuction intersects x-axis at x=7 and x=-11
\[(x+10)^2-81=0\] \[(x+10)^2=81\] \[x+10=\pm\sqrt{81}=\pm9\] so \(x+10=9\iff x=-1\) and \(x+10=-9\iff x=-19\)