TOP 5 things I learned in Precalculus 11

There were many important topics and lessons learned in this Precalculus 11 course. The following top 5 things were what I believed stood out from the rest. They, in turn, will benefit me when taking the next math course in grade 12 and beyond.

Positive and Negative Rational Exponents and its laws
The topic of Rational Exponents is easily in my Top 5 List. After learning the basics of Exponents in middle school and in high school, its importance is crucial in solving different types of algebra problems such as scientific notation, exponential growth, and exponential decay problems. Also, some real world applications need exponents to fully understand the scientific scales like the pH scale or the Richter scale.

In this course, to calculate rational exponents and to apply the Exponent Laws, some basic rules are required. Furthermore, using these Laws makes it easier to rewrite radicals as fractional exponents and vice versa. According to the rule below, the numerator of the fractional exponent is the power of the radicand. The denominator becomes the index of the root.

$x^\frac{m}{n}$ = $\sqrt[n]{x^m}$ = $(\sqrt[n]{x})^m$

For example,

$27^\frac{5}{3}$ = $\sqrt[3]{27^5}$ = $(\sqrt[3]{27})^5$

The following are examples of evaluating expressions with rational exponents:
A)

B)
Rational Expressions
A rational expression is a fraction where the numerator and denominator are polynomials. During that week of learning and understanding this concept, I was intrigued about it. Such expressions can represent real-life situations and find answers to distance-speed-time questions as well as multi-person work problems. That is why I included Rational Expressions into my Top 5.

Let’s simplify the following expression (from question 13 on page 569):

Let’s solve a different kind of rational expression (from question 10.B on page 585):

$\frac{16}{g^2+2g-12}$ = $\frac{6}{g^2-9}$

Factor each denominator and find the non-permissible values.

$\frac{16}{2(g-2)(g+3)}$ = $\frac{6}{(g-3)(g+3)}$

The non-permissible values are g=2,-3,and 3,-3.

Cross-multiply and solve.

Since -3 is a non-permissible value (as mentioned above), the only solution is 6.
Methods of solving the quadratic equation: factoring, completing the square, and quadratic formula
Factoring, the Completing the Square method, and the Quadratic Formula are used throughout this course. Understanding and mastering each method takes some time but it is worthwhile. I chose these methods of solving the quadratic equation to be in the Top 5 because these basic methods will help me with more difficult quadratic equations in future math and university math courses.

In review, the quadratic function is used in several fields of science and in daily life directly and indirectly. For example, these quadratic functions are used to calculate business profit, to determine the area of a construction site, pieces of land, or buildings, to analyze an athlete’s gameplay in sports, to calculate the speed of moving objects (for military and law enforcement), and to be applied in many fields of engineering such as automotive, aerospace, chemical, and electrical.

Every quadratic function is written in the standard form:

$ax^2 + bx + c = 0$ , where a, b, and c are constants and a ≠ 0.

When the equation contains multiple quadratic terms (for example, $x^2$ and $x$ ), it cannot be solved by isolating the variable. One strategy is Factoring the trinomial using the Zero Product Law. The law states that if the product of 2 numbers is 0, either number or both numbers equal 0.

A perfect example of factoring involving the Zero Product Law is question 7.A on page 210:

Another method, Completing the Square on a quadratic equation, is a bit trickier. Let’s try it on this example:

$x^2 + 6x = 16$

This method needs the constant term (in this case, 16) to always be on the right side of the equation (after the equal sign). In this case, we can leave it as is.

To find a “perfect square” trinomial on the left side of the equation, we must divide the coefficient of the second term $(6x)$ by 2 and then square that value.

$(\frac{6}{2})^2 = 9$

Write this value on both sides of the equation. Now, we can factor the perfect square on the left side and add up all values on the right side.

Completing the Squares gives us the roots: $x$ = -8 and 2.

Solving a quadratic equation using the Quadratic Formula is simpler but the equation must always be in the form of:

$ax^2 + bx + c = 0$ , where a, b, and c are constants and a ≠ 0.

Let’s start with an example where the quadratic terms are mixed:

Once the equation is in this particular order, we can clearly see where each value goes into the Quadratic Formula.

In this example, since $\sqrt{-71}$ is not a real number, this equation has no real roots (answers).

In addition, we took an in-depth look into the quadratic formula and learned that the expression inside the radicand of the quadratic formula is called the Discriminant. By using the Discriminant, we can determine the number of solutions (or roots) of the quadratic equation without solving the equation.

As a rule of thumb, the quadratic equation has:
Two real roots when $b^2 - 4ac \; \textgreater \; 0$
Exactly one real root when $b^2 - 4ac = 0$
No real roots when $b^2 - 4ac \; \textless\; 0$

Let’s try question 7A from page 252 to determine if this equation has one, two, or no real roots:

$2x^2 - 9x + 4 = 0$

$b^2 - 4ac$ the Discriminant

$= (-9)^2 - 4(2)(4)$

$= 81 - 32 = 49$

Because the value of the Discriminant is greater than 0, then this equation has 2 real roots.
Characteristics of the graph of a quadratic function
I believe that the parabola graph of a quadratic function tells “the whole story” in a very subtle way. This graph holds so much information that no one realizes at a first glance. This is why the Characteristics of the Graph of a Quadratic Function is in my Top 5. There are many real world examples that are modelled by parabola graphs such as throwing a sporting ball, shooting a cannon, diving from a platform, and also, hitting a golf ball. When throwing a football, the path is similar to the graph of a parabola. Analyzing that graph can answer questions such as “when does the football reach its maximum height?” and “how high does the football get?”.

The quadratic function has many useful characteristics of the graph such as the coordinates of the vertex, the domain, the range, the direction of the opening, the equation of the axis of symmetry, and the intercepts by applying its equation in the “standard form”.

$y = a(x - p)^2 + q$

From question 8.a from page 306, we will extract certain pieces of information from the equation:

$y = 3(x - 2)^2 + 1$

In this example, $p = 2$ and $q = 1$ ; therefore, the coordinates of the vertex are (2, 1).

The ( $a$ ) coefficient before the brackets is a positive 3 ; therefore, the parabola graph opens up.
The domain is as follows: x ∈ R. Also, since the graph is opens up, the vertex is a minimum point with a y-coordinate of 1.

So, that means the range is: $y \geq 1$ , y ∈ R.

In general, the equation of the axis of symmetry is x=p ; therefore, the equation should be $x = 2$ .

To find the y-intercept, substitute x=0 in the original equation and solve for $y$ .

$y = 3(x-2)^2 + 1$

$y = 3((0)-2)^2 + 1$

$y = 3(-2)^2 + 1$

$y = 12 + 1$

$y = 13$

So, the y-intercept is 13.

Finally, to find the x-intercepts, substitute y=0 in the original equation and solve for $x$ .

This equation has no solution because the radicand $(\frac{-1}{3})$ cannot be negative; therefore, there are no x-intercepts.

By looking at the completed graph, the vertex, and the x- and y-coordinates, the reversal can be done as well. In other words, the graph can be written back to its equation in factored form, General Form, or Standard Form.

In addition, another characteristic we learned to solve was coefficient, which represents the “stretch” of the graph. To find that coefficient, we must use the Standard Form of the equation and insert the coordinates of the vertex $(p, q)$ and coordinates of the second point $(x, y)$ on the graph.

$y = a(x - p)^2 + q$
Trigonometry
From middle school to high school, “SOH CAH TOA” will always be remembered as the trigonometric ratios for SINE, COSINE, and TANGENT. I chose Trigonometry as one of the Top 5 because learning this basic foundation will probably help me understand the complex features of trigonometry in grade 12 and in university. As a review, the acronym “SOH CAH TOA” is as follows:

sin θ = $\frac{Opposite}{Hypotenuse}$

cos θ = $\frac{Adjacent}{Hypotenuse}$

tan θ = $\frac{Opposite}{Adjacent}$

In addition, these same primary trigonometric ratios are defined differently on an x- and y-coordinate grid graph but the results are the same.

sin θ = $\frac{y}{r}$ cos θ = $\frac{x}{r}$ tan θ = $\frac{y}{x}$

where $r$ = $\sqrt{x^2+y^2}$

Now, take a step even further in the trigonometric ratios and we get the SINE LAWS and the COSINE LAWS (with the help from the Pythagorean Theorem) as seen below.

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ and $\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$

$c^2=a^2+b^2-2ab \cdot\cos(C)$

A word problem example involving both Sine and Cosine Laws is as follows:

Part of a roof panel measured 10m is to be installed at an angle of 49° at the base which is 15m long. The entire roof forms a triangle. Determine the measures of the other 2 angles and the length of the other side of the roof.

Using the information given, use the Cosine Law first to calculate the length of AC.

The length of AC (other side of the roof) is approx. 11.3 m.
With more information, we can now use the Sine Law to calculate the angles ∠A and ∠C.

$\frac{sin A}{a} = \frac{sin B}{b} = \frac{sin C}{c}$

Insert the known values, then solve.

$\frac{sin A}{a} = \frac{sin B}{b}$

$\frac{sin A}{15} = \frac{sin 49^{\circ}}{11.321}$

$sin A = \frac{sin 49^{\circ}}{11.321} \times 15$

$\angle A = sin^{-1} (\frac{sin 49^{\circ}}{11.321} \times 15)$

$\angle A = 89.19 = 89^{\circ}$

We can now use the angle summation in a triangle to find the other angle.

$\angle C = 180^{\circ} - (89^{\circ} + 49^{\circ})$

$\angle C = 42^{\circ}$

In summary, the other side of the roof is approximately 11.3 m and is angled 42° at the base and is angled 89° with the other roof panel.

TOP 5 things I learned in Precalculus 11

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TOP 5 things I learned in Precalculus 11

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