Answer
The tangent is horizontal at the point $(-3,0)$
Work Step by Step
$y = [ln(x+4)]^2$
$y' = 2[\frac{ln(x+4)}{x+4}]$
If the tangent is horizontal, then the slope of the tangent is 0.
We can find the x-coordinate where $y'=0$:
$y' = 2[\frac{ln(x+4)}{x+4}] = 0$
$ln(x+4) = 0$
$e^{ln(x+4)} = e^0$
$x+4 = 1$
$x = -3$
We can find the y-coordinate when $x=-3$:
$y = [ln(x+4)]^2$
$y = [ln((-3)+4)]^2$
$y = [ln(1)]^2$
$y = (0)^2$
$y = 0$
The tangent is horizontal at the point $(-3,0)$
