The price of a certain stock, in dollars, can be modeled by the function f(x)=0.75x^2-6x+20, where x equals the number of months that the stock is on the exchange. Over the same 6-month period, the price of another stock, in dollars, increases according to the function g(x)=2.75+1.50. graph the functions in the same coordinate plane to determine when f(x)=g(x). How many times during the 6-month period will both stocks be at the same price?

Question

1)three times
2)once
3)never
4)twice

Answer 1

First, let's clarify the functions given in the problem.

1. The price of the first stock is modeled by the function: \[ f(x) = 0.75x^2 - 6x + 20 \]

2. The price of the second stock is given as: \[ g(x) = 2.75 + 1.50 \] This simplifies to: \[ g(x) = 4.25 \]

We need to find out when \( f(x) = g(x) \), which translates to solving the equation: \[ 0.75x^2 - 6x + 20 = 4.25 \]

Subtract \( 4.25 \) from both sides: \[ 0.75x^2 - 6x + 20 - 4.25 = 0 \] \[ 0.75x^2 - 6x + 15.75 = 0 \]

Next, we can multiply the entire equation by 4 to eliminate the decimal: \[ 3x^2 - 24x + 63 = 0 \]

Now let's solve this quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 3 \), \( b = -24 \), and \( c = 63 \).

First, calculate the discriminant \( b^2 - 4ac \): \[ b^2 = (-24)^2 = 576 \] \[ 4ac = 4 \cdot 3 \cdot 63 = 756 \] \[ b^2 - 4ac = 576 - 756 = -180 \]

Since the discriminant is negative (\(-180\)), this indicates that the quadratic equation has no real solutions.

As a result, the graphs of \( f(x) \) and \( g(x) \) do not intersect.

Therefore, the answer to how many times during the 6-month period will both stocks be at the same price is: 3) never.