Question

Determine where f(x) intersects the x-axis. f(x)=(x+10)2−81

Answer 1

the 2 is supposed to me an exponant...

Answer 2

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

Answer 3

\[(x+10)^2-81=0\]
\[(x+10)^2=81\]
\[x+10=\pm\sqrt{81}=\pm9\] so \(x+10=9\iff x=-1\) and \(x+10=-9\iff x=-19\)

Answer 4

Thank you!