the quadratic equation $y = -6x^{2}+100x - 180$ models the stores daily profit, $y$, for selling soccer…

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.

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.

Answer

Explanation:

Step1: Set the two equations equal

Since at the intersection points, the $y$-values are equal, we set $-6x^{2}+100x - 180=-4x^{2}+80x - 150$.

Step2: Rearrange the equation

Move all terms to one - side: [ \begin{align*} -6x^{2}+100x - 180-(-4x^{2}+80x - 150)&=0\ -6x^{2}+100x - 180 + 4x^{2}-80x + 150&=0\ (-6x^{2}+4x^{2})+(100x-80x)+(-180 + 150)&=0\ -2x^{2}+20x - 30&=0 \end{align*} ] Divide through by $-2$ to simplify: $x^{2}-10x + 15=0$.

Step3: Use the quadratic formula

The quadratic formula for $ax^{2}+bx + c = 0$ is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 1$, $b=-10$, and $c = 15$. [ \begin{align*} x&=\frac{-(-10)\pm\sqrt{(-10)^{2}-4\times1\times15}}{2\times1}\ &=\frac{10\pm\sqrt{100 - 60}}{2}\ &=\frac{10\pm\sqrt{40}}{2}\ &=\frac{10\pm2\sqrt{10}}{2}\ &=5\pm\sqrt{10} \end{align*} ] When $x = 5+\sqrt{10}\approx5 + 3.162=8.162$, $y=-4(5 + \sqrt{10})^{2}+80(5+\sqrt{10})-150$. When $x = 5-\sqrt{10}\approx5 - 3.162 = 1.838$, $y=-4(5-\sqrt{10})^{2}+80(5-\sqrt{10})-150$. The intersection points $(x,y)$ represent the prices $x$ at which the daily profit from selling soccer balls is the same as the daily profit from selling footballs, and $y$ is that common profit amount.

Answer:

The $x$-values of the intersection points are $x = 5+\sqrt{10}\approx8.16$ and $x = 5-\sqrt{10}\approx1.84$. The $y$-values can be found by substituting these $x$-values into either of the original profit - equations. The intersection points represent the prices at which the daily profit from selling soccer balls is equal to the daily profit from selling footballs, and the $y$-value at the intersection points is that equal profit amount.