Exercise 67 Page 316

Multiplying $10\text{ ft}$ by this conversion factor will convert it to inches. To estimate the weight of an alligator that is $120$ inches long, we can use the trace feature of our calculator. Having drawn the graph, we press $\boxed{\text{TRACE}}.$ Then, we use the right and left arrow keys to move along the graph to get $y=120.$ Let's do it!

Therefore, the weight of the alligator is about $281$ pounds.

Next, we will use the right and left arrow keys to move along the curve and determine LeftBound?. The left bound must be below the $x\text{-}$axis. Then, we press $\boxed{\text{ENTER}}.$

Next, we will determine RightBound? proceeding in the same way. The right bound must be above the $x\text{-}$axis. Then, we again press $\boxed{\text{ENTER}}.$

Now, we will press $\boxed{\text{ENTER}}$ one more time to determine the zero of the function.

The $x\text{-}$intercept of the function is $(3.35,0).$ The $x\text{-}$intercept tells us that an alligator with a weight of $3.35$ pounds has no length. However, if an object have a weight, it must have a length. Therefore, the $x\text{-}$value does not make sense in this content.