Question

College algebra homework review... Having issues calculating this by hand and on TI-84 receiving errors like "8e12" when trying to calculate the actual quadratic equation it calls for in question B.... Please help

Answer 1

checking the vertex of this upside-down parabola, it has a vertex at (1000, 2000000), so that's the U-turn, when as the price "p" increases the revenue goes down.

`\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{a}x^2\stackrel{\stackrel{b}{\downarrow }}{+b}x\stackrel{\stackrel{c}{\downarrow }}{+c} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)`

now, if we solve the quadratic using the value of 500000

`\bf \stackrel{R(p)}{500000}=-2p^2+4000p\implies 250000=-p^2+2000p \\\\\\ p^2-2000p+250000=0`

and we run the quadratic formula on it, we get the values of x = 133.97 and x = 1866.03, one value is obviously when going upwards, the first one, and the other is when going downwards.

so we know that the R(p) is 500,000 at x = 133.97, and it keeps on going up, up to the vertex above at x = 1000, so we can say from x = [134, 1000] R(p) > 500000.