# 5.1 Quadratic functions  (Page 5/15)

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## Finding maximum revenue

The unit price of an item affects its supply and demand. That is, if the unit price goes up, the demand for the item will usually decrease. For example, a local newspaper currently has 84,000 subscribers at a quarterly charge of $30. Market research has suggested that if the owners raise the price to$32, they would lose 5,000 subscribers. Assuming that subscriptions are linearly related to the price, what price should the newspaper charge for a quarterly subscription to maximize their revenue?

Revenue is the amount of money a company brings in. In this case, the revenue can be found by multiplying the price per subscription times the number of subscribers, or quantity. We can introduce variables, $\text{\hspace{0.17em}}p\text{\hspace{0.17em}}$ for price per subscription and $\text{\hspace{0.17em}}Q\text{\hspace{0.17em}}$ for quantity, giving us the equation $\text{\hspace{0.17em}}\text{Revenue}=pQ.$

Because the number of subscribers changes with the price, we need to find a relationship between the variables. We know that currently $\text{\hspace{0.17em}}p=30\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}Q=84,000.\text{\hspace{0.17em}}$ We also know that if the price rises to $32, the newspaper would lose 5,000 subscribers, giving a second pair of values, $\text{\hspace{0.17em}}p=32\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}Q=79,000.\text{\hspace{0.17em}}$ From this we can find a linear equation relating the two quantities. The slope will be $\begin{array}{ccc}\hfill m& =& \frac{79,000-84,000}{32-30}\hfill \\ & =& \frac{-5,000}{2}\hfill \\ & =& -2,500\hfill \end{array}$ This tells us the paper will lose 2,500 subscribers for each dollar they raise the price. We can then solve for the y -intercept. This gives us the linear equation $\text{\hspace{0.17em}}Q=-2,500p+159,000\text{\hspace{0.17em}}$ relating cost and subscribers. We now return to our revenue equation. $\begin{array}{ccc}\hfill \mathrm{Revenue}& =& pQ\hfill \\ \hfill \mathrm{Revenue}& =& p\left(-2,500p+159,000\right)\hfill \\ \hfill \mathrm{Revenue}& =& -2,500{p}^{2}+159,000p\hfill \end{array}$ We now have a quadratic function for revenue as a function of the subscription charge. To find the price that will maximize revenue for the newspaper, we can find the vertex. $\begin{array}{ccc}\hfill h& =& -\frac{159,000}{2\left(-2,500\right)}\hfill \\ & =& 31.8\hfill \end{array}$ The model tells us that the maximum revenue will occur if the newspaper charges$31.80 for a subscription. To find what the maximum revenue is, we evaluate the revenue function.

$\begin{array}{ccc}\hfill \text{maximum revenue}& =& -2,500{\left(31.8\right)}^{2}+159,000\left(31.8\right)\hfill \\ & =& 2,528,100\hfill \end{array}$

## Finding the x - and y -intercepts of a quadratic function

Much as we did in the application problems above, we also need to find intercepts of quadratic equations for graphing parabolas. Recall that we find the $\text{\hspace{0.17em}}y\text{-}$ intercept of a quadratic by evaluating the function at an input of zero, and we find the $\text{\hspace{0.17em}}x\text{-}$ intercepts at locations where the output is zero. Notice in [link] that the number of $\text{\hspace{0.17em}}x\text{-}$ intercepts can vary depending upon the location of the graph.

Given a quadratic function $\text{\hspace{0.17em}}f\left(x\right),\text{\hspace{0.17em}}$ find the $\text{\hspace{0.17em}}y\text{-}$ and x -intercepts.

1. Evaluate $\text{\hspace{0.17em}}f\left(0\right)\text{\hspace{0.17em}}$ to find the y -intercept.
2. Solve the quadratic equation $\text{\hspace{0.17em}}f\left(x\right)=0\text{\hspace{0.17em}}$ to find the x -intercepts.

## Finding the y - and x -intercepts of a parabola

Find the y - and x -intercepts of the quadratic $\text{\hspace{0.17em}}f\left(x\right)=3{x}^{2}+5x-2.$

We find the y -intercept by evaluating $\text{\hspace{0.17em}}f\left(0\right).$

$\begin{array}{ccc}\hfill f\left(0\right)& =& 3{\left(0\right)}^{2}+5\left(0\right)-2\hfill \\ & =& -2\hfill \end{array}$

So the y -intercept is at $\text{\hspace{0.17em}}\left(0,-2\right).$

For the x -intercepts, we find all solutions of $\text{\hspace{0.17em}}f\left(x\right)=0.$

$0=3{x}^{2}+5x-2$

In this case, the quadratic can be factored easily, providing the simplest method for solution.

$0=\left(3x-1\right)\left(x+2\right)$
$\begin{array}{cccccc}\hfill h& =& -\frac{b}{2a}& \hfill \phantom{\rule{2em}{0ex}}k& =& f\left(-1\right)\hfill \\ & =& -\frac{4}{2\left(2\right)}\hfill & & =& \hfill 2{\left(-1\right)}^{2}+4\left(-1\right)-4\\ & =& -1\hfill & & =& -6\hfill \end{array}$

So the x -intercepts are at $\text{\hspace{0.17em}}\left(\frac{1}{3},0\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(-2,0\right).$

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