## How Do You Find Special Points on a Parabola?

Let’s look at how to use formulas for a parabola to get certain important points on a parabola.

Problem For the parabola y = 2x2 + 3x – 2, locate the points below.

a. The y-intercept.

Solution At the y-intercept, the x value is zero. This means that we need to set x = 0 in the equation:

y = 2(02) + 3(0) – 2 = -2

Putting this together, the y-intercept is at (0, -2).

b. The vertex.

Solution The vertex is located using the formula   where the values of a, b, and c come from the equation. In this case, a = 2, b = 3, and c = -2. This gives an x value on the intercept of

To find the corresponding y value, put this value into the equation,

This means the vertex is at (-3/4, –25/8).

c. The x-intercepts.

Solution At the x-intercepts, the y value is zero. Putting this into the equation yields

0 = 2x2 + 3x – 2

This equation is solved with the quadratic formula,

Put the values from the equation (a = 2, b = 3, and c = -2),

The x intercepts are at (-2, 0) and (1/2, 0).

All of these points are shown in the graph of the parabola below.

## What Does It Mean To Scale A Variable?

On many problems, the most challenging aspect is keeping track of the variables and what they represent. When the variables are scaled in hundreds, thousands, or even millions, we need to pay careful attention.

Scaling means that the quantity has been divided by some amount. For instance, we can write the amount $14,400,000 as 14.4 million dollars. To scale$14,400,000 in millions, divide by 1,000,000 and then add the word “million” after the result. We could also scale in thousands by dividing by 1000. This gives 14,400 thousand dollars.

Let’s look at an application with scaling.

## How Does Function Notation Work?

In calculus, we will need to  take a function (x) and write out (x+h) for that function. Let’s look at how to do this properly.

To do a problem like this, you need to understand exactly what the x in (x) represents and what the f represents. Let’s look at the function (x) = x2x. A function is a process. In this case, it is the process of

1. Square the input
2. Take the result and subtract the input

Notice that there is no mention of the x in the formula. That is because it is a placeholder representing the input. There is nothing special about x. We could have just as easily used a different letter as a placeholder for the input. If I had wanted to call the input t, I would have written

(t) = t2 – t

If the input had been represented by the word dog, I would have written

(dog) = dog2 – dog

The input variable is simply a placeholder…if a number is put in its place like 7, we get

(7) = 72 – 7 = 42

Notice that the process is the same. Square the input and subtract the input from the result. In this case, the input is 7 so we are squaring 7 and then subtracting 7 from the result.

Many students are confused by f(x+h). Now the input is represented by x+h instead of x. This means we need to square it and then subtract x+h from the result.

(x+h) = (x+h)2 – (x+h)

We can simplify this by foiling out the square,

(x+h) = (x+h)(x+h) = x2 +2xh + h2

And removing the parentheses after the subtraction we get

(x+h) =  x2 +2xh + h2xh

The handout below has more examples with this function.

## How Do You Find Break-Even Points?

If you are looking for where the break-even points are, you must determine the quantity for which

revenue = cost

Alternately, you may find the profit by calculating

profit = revenue – cost

The problem below demonstrate these strategies starting from a demand function and a cost function. To apply either of the relationships above, you need to form the revenue function from

revenue = (price)(quantity)

where the price is given by the demand function and Q represents the quantity.

Problem The demand function for Q units of a product is given by

$latex \displaystyle D\left( Q \right)=16-1.25Q$

The cost function is given by the function

$latex \displaystyle C\left( Q \right)=2Q+15$

a. Find the revenue function R(Q).

b. Find the break-even point(s)?

c. On a graph of R(Q) and  C(Q), where do the break-even points lie?

d. Find the profit function P(Q).

e. Where do the break-even points lie on the graph of P(Q)?

Solution 1 To find the break-even point, this group of students set R(Q) = C(Q). This results in a quadratic equation. They moved all terms to one side and used the quadratic formula to find the quantities at which the revenue is equal to the cost.

Solution 2 This group of students found the profit function P(Q) first. Then they set it equal to zero to find the break-even points. Like the first solution, they also needed to use the quadratic formula.

Both techniques lead to the same break-even points and are equally valid. The only thing the second solution left out was the graph of the profit function showing the break-even points at the zeros (horizontal intercepts) of the function.

## How Do You Interpret Inputs and Outputs for a Model?

In the example below, we want to look at the inputs and outputs for a function and interpret what they tells us. In both examples, the function is a quadratic function that models the rise and fall of an object thrown in the air.

Example 1 Suppose a ball is thrown into the air has its height (in feet) given by the function

$latex \displaystyle h(t)=6+128t-16{{t}^{2}}$

a. Find h(1) and explain what it means.

b. Find the height of the ball 4 seconds after it is thrown.

c. Test other values of  to decide if the ball eventually falls. When does the ball stop climbing?

Example 2 Suppose a ball is thrown into the air has its height (in feet) given by the function

$latex \displaystyle h(t)=6+96t-16{{t}^{2}}$

a. Find h(1) and explain what it means.

b. Find the height of the ball 3 seconds after it is thrown.

c. Test other values of  to decide if the ball eventually falls. When does the ball stop climbing?

The key thing is to test the function at enough points to convince yourself that the peak is really the peak. If the peak height occurs at x = 4.2, will finding h(3), h(4), and h(5) be enough to find that peak? If we have a graph, we can use it to find the peak. But if we only have the function we need to fine tune the input to zoom in on wherever the peak is.

In each case, the values of t are in seconds and h(t) is in feet. We want to find an h(t) value that is higher then those on either side.