SOLUTION: A vase 25cm tall is positioned on a bench near a wall. The shape of the vase follows the curve y=(x-10)^2, where y is the height of the vase and x is the distance of the vase from
Question 1092468: A vase 25cm tall is positioned on a bench near a wall. The shape of the vase follows the curve y=(x-10)^2, where y is the height of the vase and x 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.
There is a diagram on the question and it is a parabola that shows nothing except the turning point is on 0 on the x axis. and there is a small gap between the wall and the top of the parabola.
I really need help answering these questions because I have a test and I don't know anything! Please write all of your workings so I understand! Thank you Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! let's look at the graph of y=(x-10)^2
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Note x-axis is the distance of the vase from the wall and y-axis is the
height of the vase
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a) 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
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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
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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
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d) y = (x-10+5)^2
y = (x-5)^2
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