Question

the quadratic equation $y = -6x^2 + 100x - 180$ models the stores daily profit, $y$, for selling soccer balls at $x$ dollars. the quadratic equation $y = -4x^2 + 80x - 150$ models the stores 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.

Explanation:

Step1: Set the equations equal

To find the intersection points, set \(-6x^{2}+100x - 180=-4x^{2}+80x - 150\)

Step2: Simplify the equation

Bring all terms to one side: \(-6x^{2}+100x - 180 + 4x^{2}-80x + 150 = 0\)
Simplify: \(-2x^{2}+20x - 30 = 0\)
Divide by -2: \(x^{2}-10x + 15 = 0\)

Step3: Solve the quadratic equation

Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c = 0\)
Here, \(a = 1\), \(b=-10\), \(c = 15\)
\(x=\frac{10\pm\sqrt{100 - 60}}{2}=\frac{10\pm\sqrt{40}}{2}=\frac{10\pm2\sqrt{10}}{2}=5\pm\sqrt{10}\)
Approximate values: \(x\approx5 + 3.16 = 8.16\) or \(x\approx5 - 3.16 = 1.84\)

Step4: Find corresponding y - values

For \(x = 5+\sqrt{10}\approx8.16\)
Using \(y=-4x^{2}+80x - 150\)
\(y=-4(8.16)^{2}+80(8.16)-150\approx - 4(66.59)+652.8 - 150\approx - 266.36+652.8 - 150\approx236.44\)
For \(x = 5 - \sqrt{10}\approx1.84\)
\(y=-4(1.84)^{2}+80(1.84)-150\approx - 4(3.39)+147.2 - 150\approx - 13.56+147.2 - 150\approx - 16.36\) (We can also use the other equation, the result will be the same)

for context:
The intersection points \((x,y)\) mean that when the price of the soccer ball and the football is \(x\) dollars, the daily profit \(y\) from selling soccer balls and footballs is the same. For example, when \(x\approx1.84\) dollars, both have a profit of approximately \(- 16.36\) dollars (a loss), and when \(x\approx8.16\) dollars, both have a profit of approximately \(236.44\) dollars.

Answer:

The intersection points are approximately \((1.84, - 16.36)\) and \((8.16,236.44)\) (or \((5 - \sqrt{10},-4(5 - \sqrt{10})^{2}+80(5 - \sqrt{10})-150)\) and \((5+\sqrt{10},-4(5 + \sqrt{10})^{2}+80(5 + \sqrt{10})-150)\)). In context, these points represent the prices (\(x\) - values) at which the daily profit from selling soccer balls and footballs is equal, with the corresponding profit values (\(y\) - values).