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.

Continue reading “What Does It Mean To Scale A Variable?”

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 Graph An Absolute Value Function in Desmos and WolframAlpha?

Graphing an absolute value function can be a bit deceiving. Depending on the technology you use, the graph you get may not actually represent the function well.

Let’s graph the function

using WolframAlpha and Desmos. These two online graphing tools are both free to use and can produce excellent graphs.

To graph this function in WolframAlpha, go to the website and type this in the box on the screen.

Both the numerator and denominator need to be in parentheses. The absolute value function in WolframAlpha is “abs”. Putting this in front of (x+4) means the absolute value of the quantity x + 4.

Press return to give the following result.

The graph consists of a horizontal section at y = -1 and another at y = 1. These sections are connected by a vertical line at x = -4. This is problematic since this is not a function…it does not pass the vertical line test at x = -4.

Let’s try graphing this function in Desmos.

As shown in the video above, the graph of this function looks like this in Desmos.

This looks similar to the WolframAlpha version, except that the tow horizontal pieces are not connected. As noted in the video,

is undefined at x = -4. This is because x = -4 causes the denominator to be zero.

This might not seem like a big deal. But if you were determining whether the function was continuous at x = -4, the two graphs would lead to different conclusions. The WolframAlpha graph would lead you to think the function is continuous. Desmos would give the opposite conclusion.

In this case, Desmos gives a more accurate graph since it shows the discontinuity at x = -4. An even better version of this graph would be to include open circles at x = -4.

This not only shows the discontinuity, but also indicates that the function is undefined at x = -4. To put these on the graph I downloaded the image and then added the circles in an image editing program like Paint.