Question

A vase can be modeled using x squared over 6 and twenty five hundredths minus quantity y minus 4 end quantity squared over 42 and 25 hundredths equals 1 and the x-axis, for 0 ≤ y ≤ 20, where the measurements are in inches. Using the graph, what is the distance across the base of the vase, and how does it relate to the hyperbola?

Answer 1

Before we start, let's plot the graph of the hyperbola:

The problem wants the distance across the base of the base, we can see that at y = 0 (base) the "vases" touches x = 2.935. that's the distance between the center of the base and the end of it, looking at the image we can see that basically, it's the distance between the x-intercept, we know that half of the distance is 2.935, then the distance is

`\begin{gathered} (d)/(2)=2.935 \\ \\ d=2\cdot2.935 \\ \\ d=5.87\text{ inches} \end{gathered}`

Therefore, the correct answer is 5.87 inches, distance between x-intercept