Parabolic Problems

Another problem I have seen lately is that of parabolas and symmetry, as well as that of intervals, domains, ranges, and other graphical dilemmas.

If one is asked to define the interval over which a function is increasing, decreasing, or remaining the same, the answer is always (x₁, x₂). One never uses brackets when referring to an interval! However, if one is asked to define a domain or range, the answer could include one or more brackets: (x₁, x₂], [x₁, x₂), or [x₁, x₂] if referring to the domain; (y₁, y₂], [y₁, y₂), or [y₁, y₂] if referring to the range. (x₁, x₂) and (y₁, y₂) can also be used to describe the domain and range, respectively, in some cases.

A function is said to be increasing if the y-values are getting larger, decreasing if they are getting smaller, and constant if they are remaining steady. The domain of the function is the entire group of all the x-values for which the function is continuous, while the range is the entire group of y-values for which the function is continuous.

The bracket is used when the function is continuous over the requested domain or range, but in the case of infinity, the parenthesis is used, instead. If the function is discontinuous, the parenthesis is used.

y = x² + 6x + 7

Given the equation y = x² + 6x + 7, one can not immediately deduce what the function looks like, except that it should be a parabola.

This is where the vertex form of the parabola comes in:
y = a(x - h)² + k.
To get the equation in this form requires a little bit of thought. We need the original equation in the

form of a perfect polynomic square—in this case y = x² + 6x + 9.

Start with the original equation: y = x² + 6x + 7. Add parentheses: y = (x² + 6x + 7). In this case, add 2 inside the parentheses, and subtract 2 outside the parentheses: y = (x² + 6x + 7 + 2) - 2. This allows the function to have the same value as it had originally, while also putting it in a more convenient form. Next, factor the function: y = (x² + 6x + 9) - 2—y = (x + 3)² - 2. This—y = (x + 3)² - 2—is the final "vertex form" of the equation; we can now find the vertex, see whether we have a positive or negative parabola, which will also tell one whether there is a maximum point or a minimum point.

Looking at the equation: y = (x + 3)² - 2, there is no "sign" before the parenthesis—y = _(x + 3)² - 2, meaning this is a positive, upward-facing parabola, with the vertex at the minimum. Moving on to the next item of interest—y = (x + 3)² - 2—the x-value of the vertex is always the opposite of the number in the parentheses; in this case, x = -3. And lastly, the equation can give us the y-value of the vertex: y = (x + 3)² - 2; the y-value of the vertex is always equal to the number to the right of the parentheses—if there is no number there, y = 0. However, in this case, y = -2. This gives a resultant vertex of (-3, -2)

The x-value can also be found using the equation: x = - ^b⁄_2a with the quadratic formula.

If we had had y = x² - 6x + 7, instead, the function would have simplified to y = (x - 3)² - 2, with a positive, upward-facing parabola, with the vertex again at the minimum—this time at (3, -2), however.

If a question asks about symmetry, the equation of the line of symmetry in a vertical parabola—as seen in the example—is equal to the x-value of the vertex. In the example above, the axis of symmetry is found at x = -3. For a horizontal parabola, the axis of symmetry would be found at the y-value of the vertex.

Questions asking about whether a function is symmetrical about the x-axis, y-axis, or origin, are generally simple as well. If the function is mirrored, or looks the same, on both sides of either the x-axis or y-axis, it is symmetrical. To be symmetrical about the origin, however, the function must look the same across the origin diagonally. If it looks one way in Quadrant I, it must be mirrored across, flipped, into Quadrant III; not either of Quadrants II or IV next to it—it has to be the quadrant directly across the origin—diagonally.

If you are ever needing an online calculator, maybe one that's more helpful than just typing your question into the search bar, I have found Mathway to be quite useful when I don't have my calculator with me!

I hope this post is helpful to those who may be struggling with this—especially with midterms approaching, and for some, almost over; however, if any clarification would be of assistance, please let me know via my contact page found here: Contact Me. Thanks for reading! If this was of assistance to you, please share, follow me, or let me know through the comments section below.