Question

The quadratic equation y = –6x2 + 100x – 180 models the store’s daily profit, y, for selling soccer balls at x dollars. The quadratic equation y = –4x2 + 80x – 150 models the store’s daily profit, y, for selling footballs at x dollars. Use a graphing calculator to find the intersection point(s) of the graphs, and explain what they mean in the context of the problem

Answer 1

Sample response:

points of intersection represent when the price and profit are the same for each type of ball.

The intersection points are approximately

(8.16, 236.49) and (1.84, –16.49).

When the store charges $8.16 for each type of ball, they make the same profit from each ball, approximately $236.49.

Charging $1.88 provides no profit for either type of ball.

Answer 2

Answer:

Step-by-step explanation:

The quadratic equation    $y = -6x^2 + 100x - 180$        models the store’s daily profit, y, for selling soccer balls at x dollars.

The quadratic equation    $y = -4x^2 + 80x - 150$    models the store’s daily profit, y, for selling footballs at x dollars.

Now we are supposed to find the intersection point(s) of the graphs using graphing calculator

Refer the attached graph.

$y = -6x^2 + 100x - 180$   -- Green line

$y = -4x^2 + 80x - 150$    -- Purple line

So, from the graph we can see there are two intersection points :

(8.333,236.667)

(2.053,0)

So, from solution (8.333,236.667) we depict that at x = 8.333 dollars the store's daily profit y = 236.667 for selling football or soccer ball.

From solution (2.053,0)we depict that at x = 0 dollars the store's daily profit y = 2.053 for selling football or soccer ball.