The price c, in dollars per share, of a high-tech stock has fluctuated over a twelve-year period according to the equation c= 14 +12x – x2, where x is in years. the price c, in dollars per share, of a second high-tech stock has shown a steady increase during the same time period according to the relationship c = 2x + 30. for what values are the two stock prices the same?

For this case we have the following equations:
 C = 14 + 12x - x2
 C = 2x + 30
 Equating the equations we have:
 14 + 12x - x2 = 2x + 30
 Rewriting we have:
 -x2 + 10x - 16 = 0
 Solving the polynomial we have:
 x1 = 2
 x2 = 8
 
 the two stock prices are the same for:
 x1 = 2
 x2 = 8

Tim's with a y-intercept of $2,780

Step-by-step explanation:

In order to solve this problem we must first find the equation Tim is describing with words. So let's take the first part of the problem:

"The amount of money in a savings account increases at a rate of $225 per month."

this can be written as an equation like this:

where y is the amount of money in the savings account, x is the number of months and is the original ammount of money in the account.

Next, the problem states:

"After eight months, the bank account has $4,580 in it."

so we can use this value to find the initial amount of money in the account, so we plug that value in:

when solving for

we get:

so the equation can now be completed, so we get:

y=225x+2780

whose y-intercept is 2780.

Let's analyze Paul's equation now:

y-1400=56(x+26)

so let's rewrite this equation in slope intercept form. We can do that by solving the equation for y, so we get:

y=56x+1456+1400

which yields:

y=56x+2856

in this case, the y-intercept is 2856.

As you may see, Tim's equation hast the smaller y-intercept which was $2780.