Question

7. A vase 25 cm tall is positioned on a bench near a wall as shown. The shape of the vase follows the curve y = (x - 10)^2, where y cm is the height of the vase and x cm is the distance of the vase from the wall.

a. How far is the base of the vase from the wall? b. What is the shortest distance from the top of the vase to the wall?
c. If the vase is moved so that the top just touches the wall, find the new distance from the wall to the base.
d. Find the new equation that follows the shape of the vase.​

Answer 1

Explanation:

the base of the vase will be where the vase touches the x-axis, that is 10 cm, therefore, the base is 10 cm from the wall

:

b) 25 = x^2 -20x +100, we solve for x to find the closest distance since as we move up the vase the distance to the wall gets closer(assume the y-axis is the wall), then

x^2 -20x +75 = 0 (x-15) * (x-5) = 0

x = 15 and x = 5

we reject x = 15

the shortest distance from the top of the vase to the wall is 5 cm

:

c) this is a left shift of the equation y = (x-10)^2

from b) we know that the left shift is 5 cm

10 - 5 = 5 cm from the wall to the base

:

d) y = (x-10+5)^2

y = (x-5)^2