$y=a(x-h)^2+k$ where h is side to side movement (it's also the x coordinate of the vertex) and k is the up or down movement (it's also the y coordinate of the vertex). If there is no up or down movement, the k value is 0. (We don't need to worry about the value for a here; it's 1 but that doesn't change anything for us in our problem). Movement to the right is positive, so we are moving +10. Filling that into our equation:

$y=(x-(+10))^2$ and simplified:

$y=(x-10)^2$ That is the parent graph shifted 10 units to the right.

8 0

2 years ago

A charter airline finds that on its Saturday flights from Philadelphia to London, all 120 seats will be sold if the ticket price

Nataliya [291]

Answer:

a) $P=3(120-s)+200$

b) $215 \leq P \leq 290$

Step-by-step explanation:

For this case we can use a linear model to solve the problem.

s) Create an equation to express the increase on the price tickets and the number of seats sold

$s$ number of seats, if w analyze the info given the number of seats after increase the price is given by $120-s$ .

And let P the price for the ticket. So after the increase in ticket price the expression for the increase is P-200.

We have an additional info, for each increase of $3 the number of setas decrease 1. And the equation that gives to us the price change in terms of the increase of price is:

$P-200=3(120-s)$

So then our linear equation is given by:

$P=3(120-s)+200$

b) Over a certain period, the number of seats sold for this flight ranged between 90 and 115. What was the corresponding range of ticket prices?

So for this case we just need to replace the limits into the linear equation and see what we got:

$P_L=3(120-90)+200=290$

$P_U=3(120-115)+200=215$

So the corresponding range of ticket prices is:

$215 \leq P \leq 290$

4 0

2 years ago