#### To determine

**To find:**the point on the curve y=[ln(x+4)]2 at which the tangent is horizontal

#### Answer

Hence the point at which the tangent is horizontal on the curve y=[ln(x+4)]2 is (−3,0)

#### Explanation

Given y=[ln(x+4)]2

Differentiating w.r.t x

y'=2ln(x+4).1x+4

y'=2ln(x+4)x+4

The tangent line is horizontal, when slope of the curve is zero

∴ln(x+4)=0⇒x+4=e0⇒x+4=1

⇒x=−3

When x=−3,y=(ln1)2=0

Hence the point is (−3,0)

**Conclusion:** Hence the point at which the tangent is horizontal on the curve y=[ln(x+4)]2 is (−3,0)