Category Archives: Math 11

Quadratic Equations in Vertex Form

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My Parabola:

Significance of Variables in My Equation

As shown above, the vertex form of a quadratic equation is: y = a(x – h)2 + k.

Within this equation, the value of ‘a’ determines whether the parabola opens upwards or downwards, as well as how steep or flat it is. A positive value for a makes the parabola open upwards, wile a negative value of a makes it open downwards. Additionally, the wideness of the parabola increases as the value of a decreases. The negative value of a in my equation clearly shows that the parabola will open downwards. And since the value of a is quite small, it will be considerably more wide in comparison to the parent function. The value of a in my parabola is -1/3.

When identifying the vertex of a parabola, the h value is half of the entire process. Our h value is crucial as it is our x coordinate for the vertex and determines the horizontal shift and the axis of symmetry for our parabola. If h is positive, the parabola will shift right and if h is negative, it will shift to the left. In our original equation, our h value is 5, allowing us to infer that our parabola will shift right, and our vertex will be (5,y). This means our vertex has shifted right from our parent function by 5 units.

The second half of finding our vertex is the k value. This is what gives us our y coordinate to our vertex. The k value dictates the vertical shift of our parabola and represents the minimum/maximum value. It represents the minimum or maximum value depending on whether the graph opens up or down. If the graph opens upwards, the vertex is at the lowest point of the parabola, and therefore has a minimum value. If the graph opens downwards, however, the vertex is at the highest point in the parabola, meaning that it represents the maximum value. With the example above our k value is -4 giving us a y coordinate to our vertex. This means our vertex has shifted down from our parent function by 4 units. This information gives us the vertex of (5, -4).

And, since the graph opens downwards (due to the negative a value), we can see that the maximum value of the parabola is -4.

Summary – Stats of the Parabola

(a) value: -1/3
(h) value: 5
(k) value: -4
Vertex: (5,-4)
Axis of Symmetry: x = 5
Minimum/Maximum value: y <= -4

Self-Assessment

1. How did you represent the same mathematical idea in multiple ways in this assignment?

I represented the mathematical idea by showing that making alterations to the a, h, and k values of the parabola can have an effect on the position and shape of the graph. I did this by explaining the significance of each variable in my own equation, as well as using Desmos to graph my function.

2. State the relevant mathematical vocabulary words you used to demonstrate your understanding.

Parabola: A curved graph that goes through points according to its quadratic function.
Vertical/Horizontal Shift: The number of units that the graph is shifted left/right and up/down
Quadratic Equation: A polynomial function with 5 variables
Vertex: The highest/lowest point of the parabola where the graph changes direction
Origin: The midpoint of the graphing paper (0,0)

3. How did you use formatting to share this information in a clear and organized way?

I used Desmos to graph both the parent function as well as my assigned function. I then edited the image by adding the corresponding equation and vertex of each graph.

Facing a Challenge

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The first time I tried this problem I found it challenging because of how complicated it seemed, due to the large number of words in the question. I consider the question to be a very interesting question, however, as there are a few different variables, and it is formatted in the form of a word problem.

When I first attempted to solve the problem, I ran into some issues when performing the task of squaring both sides, which was necessary in the removing of the radical and simplifying the question. In trying to do this step, I failed to remember to square the π, and therefore ended up with an answer drastically different from the one in the workbook.

When I was stuck on this problem, I read the question carefully in my mind and made a list consisting of the most important information which was in the problem, such as the equation and the given variables. I also used my notes to see if there was a conceptual mistake in my way of thinking. Additionally, I asked a peer how they approached the question, and considered the problem using their methods. By using these strategies while attempting the problem a second time, I was able to effectively solve for the correct solution.

While solving this problem, I needed to remember how to interpret a word problem and use the given information properly. Order of operations (Bedmas), the FOIL technique, and knowing how to remove radicals from an equation were also important concepts for solving this question. Since it was formatted as a word problem, another useful strategy I used was to cross out any irrelevant words in the wording of the question, which effectively made the problem more clear.

Next time I encounter a difficult problem, I will first consider the given information and see how it will help me solve the problem. For example, this could include any easily variables or formulas/equations, such as the one in this problem (t = 2π√L ÷g). Additionally, I will recall previous concepts which apply to the problem by looking at my notes and useful videos related to the topic if necessary.