Review Ia - HackMD

--- title: Review Ia tags: teaching, review, calculus --- # Review I (Algebra) ## Equations of lines ### Two typical forms for equations of lines 1. Slope-intercept form ::: info Slope is a number $m$ and the $y$-intercept a number $b$:$$y=mx+b$$ ::: 2. Point-slope form ::: info Slope is a number $m$ and knowing a point $(x_0,y_0)$ is on the line:$$y - y_0 = m(x - x_0)$$ ::: ### Recall how to find slope, given two points on line If $(x_1,y_1)$ and $(x_2,y_2)$ are the two points:$$m = \frac{y_1-y_2}{x_1-x_2}$$ ### Examples 1. If a line has slope $1.5$ and passes through the point $(2,-1)$, find an equation for the line. ▷ **Solution** We are told $m=1.5$ is the slope. Using point-slope form, $$y - (-1) = 1.5(x - 2).$$ This can be rewritten as $y = 1.5(x-2) - 1$; or $$y = 1.5x-4.$$ 2. If a line passes through points $(-4,3)$ and $(2,1)$, find an equation for the line. ▷ **Solution** First find the slope: $m = \frac{3-1}{-4-2} = -\frac{1}{3}$. Next, you can either find the intercept, or use point-slope form (picking either one of the two points to use as $(x_0,y_0)$). - [Approach 1](https://i.imgur.com/BlEmIdp.mp4), by finding the intercept. - [Approach 2](https://i.imgur.com/nL1eNKb.mp4), by using point-slope form. ### Some exercises for practice There are examples in Section 1.2 of the [textbook](https://openstax.org/books/calculus-volume-1/pages/1-introduction). Consider [this one](https://openstax.org/books/calculus-volume-1/pages/1-2-basic-classes-of-functions#CNX_Calc_Figure_01_02_003), or Checkpoint 1.9 just after it. You might also try (from the textbook): | Section 1.2 | Task | Exercises | | -------- | -------- | -------- | | | Write equation | 67, 70, 71, 73 | | | Find slope or intercept | 77, 81 | | | "Word" problem | 103 | For more review on lines and slopes, you might like the following two videos. > [Video 1](https://youtu.be/rpMu98yRk40) (Also covers calculating distance in the plane.) > [Video 2](https://youtu.be/MXV65i9g1Xg) <hr style="background-color:black;"/> ## Quadratic equations ### Solving quadratic equations A quadratic equation has 0,1, or 2 solutions. :::info For an equation $$ax^2+bx+c=0,$$ where $a\ne0$, then solutions must satisfy $$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$ ::: - If $b^2-4ac$ is positive, this gives two solutions. - If $b^2-4ac$ is negative, there are no solutions (in $\mathbb R$). - If $b^2-4ac=0$ then there is just one solution, $x = \frac{-b}{2a}$. *Some* quadratics can be factored (with integers), which lets you solve by setting each factor to zero. > [Video on factoring quadratics](https://youtu.be/YtN9_tCaRQc) ### Exercises 1. Solve the equation $(x-1)(x-2) = 7-x^2$. ▷ **Solution** [Found Here](https://i.imgur.com/nyMPw7H.mp4). 2. Find where the graph $y=x^2-2x-63$ intersects the $x$-axis. :::spoiler The two solutions are $x =-7$ and $x=9$. ::: ### More exercises - Checkpoint 1.10 in Section 1.2 of the textbook. - [Exercises 318 and 319](https://openstax.org/books/calculus-volume-1/pages/1-chapter-review-exercises) in the Chapter 1 review exercise. (Exercise 319 boils down to solving a quadratic equation; think about why.) <hr style="background-color:black;" /> ## Polynomial and root functions ### Polynomials A polynomial in one variable: made by adding and subtracting ["terms"](https://en.wikipedia.org/wiki/Monomial), each of which is a number times a *positive integer* power of the variable (like $3x^5$), or just a number. An example of a polynomial in the variable $t$: $$-5t^4+8.2t^3+3t-2.$$ > (Note) A polynomial *can* consist of just one term. A polynomial function of $x$ is a function that has a defining expression that is a polynomial in $x$. The domain of any polynomial function is $\mathbb R$ (or $(-\infty,\infty)$, in interval notation). > (Note) Domain being $\mathbb R$ is because any real number can be multiplied by itself some number of times, can be multiplied by other numbers, and can be added to and subtracted from other numbers. ### Degree of a polynomial, Graphs The degree of a polynomial is the largest power of the variable. The number being multiplied by this power (which might be 1, and not written) is called the leading coefficient. The degree of the polynomial has an effect on the "end behavior" of the graph of the function. :::info **Even degree** - positive leading coefficient: the graph rises (extends upward) on both the left and right ends; - negative leading coefficient: the graph falls (extends downward) on both the left and right ends. <hr style="height:1px;background-color:steelblue;"/> **Odd degree** - positive leading coefficient: the graph rises on the right end and falls on the left end; - negative leading coefficient: the graph falls on the right end and rises on the left end. ::: ![](https://i.imgur.com/jgoe7Ty.png) <div style="font-weight:bold;text-align:center;">▲ Image credit: <a href="https://math.libretexts.org/Courses/Borough_of_Manhattan_Community_College/MAT_206_Precalculus/3%3A_Polynomial_and_Rational_Functions_New/3.4%3A_Graphs_of_Polynomial_Functions">Mathematics LibreTexts, Burough of Manhattan Community College, Figure 3.4.11</a></div> More worked examples. > [Video on graphing polynomials](http://tv.hawkeslearning.com/VideoPlayerSingle.htm?PlayerID=5715214968001#) On your own, you can also try | Section 1.2 | Task | Exercises | | -------- | -------- | -------- | | Just a,b,c. | Find degree, zeros, $y$-int | 83, 84, 86, 87 | ### Root functions Root functions are those that have a (positive) fractional power of the input variable (or, of a polynomial in the variable). Two examples of root functions: $$\sqrt[5]{x^2} = x^{2/5}\ ;\qquad\qquad \sqrt{x-1} = (x-1)^{1/2}.$$ If the fraction in the power has an even number in its denominator, then a negative number cannot be raised to that power. This would exclude numbers from the domain. > Don't forget the rules of how exponents work when multiplying and/or dividing different powers of a variable $x$. For example, $$x^{2/5}x^2 = x^{\frac{2}{5}+\frac{10}{5}} = x^{12/5};$$ $$\frac{x}{\sqrt x} = x^1x^{-1/2} = x^{1/2} = \sqrt x\quad (\text{for positive } x).$$ If you've forgotten, remind yourself; here's a [video](https://www.youtube.com/watch?v=OEN9kENpvtU) (there are many others). ### Exercises 3. Find a simpler expression for $\frac{x^{11/3}}{x^{2/3}}$. ::: spoiler Simpler expression: $x^3$ ::: 4. Find the domain of $g(x) = \sqrt{2x+1}$. ::: spoiler Values of $x$ where $2x+1 < 0$ must be excluded. The solution to that inequality is $x < -1/2$, and so the domain of $g$ is $[-0.5,\infty)$. ::: 5. Find the domain of $h(x) = ((x-5)^{1/3} + x^{1/2})^2$. ::: spoiler **Answer**: The domain is $(0,\infty)$. ::: For more practice try | Section 1.1 | Task | Exercises | | -------- | -------- | -------- | | | Find the domain | 14, 15, 16, 17, 20 | <hr style="background-color:black;" /> ## Algebraic functions An algebraic function is defined by adding, subtracting, multiplying, or even dividing some number of different polynomials or root functions. So, algebraic functions may have some fractional expressions in their definition. Some examples are below. If there are no fractional exponents, then such a function can be called a rational function. To determine the domain, you must consider how the root functions that are involved will affect the domain. Also, you must exclude any values of $x$ that would make the denominator of a fractional expression equal to zero. ::: danger :warning: <span style="padding-left:1.5cm;">Dividing by zero **never** makes sense.</span> ::: ### Example 3. By simplifying, verify that, for $x\ne0$ and $x\ne\pm1$, the rational expression $\frac{1-\frac{1}{x}}{1-\frac{1}{x^2}}$ is equivalent to $\frac{x}{x+1}$. > When $x\ne0$, we can multiply by $\frac{x^2}{x^2}$ and not change the value. So, $$\frac{(1-\frac{1}{x})}{(1-\frac{1}{x^2})}\frac{x^2}{x^2} = \frac{x^2 - x}{x^2-1} = \frac{x(x-1)}{(x+1)(x-1)} = \frac{x}{x+1}.$$ > When $x=-1$ then neither expression is defined; when $x=0$ or $x=1$ then final expression is defined, but the orriginal is not. ### Exercises 6. Find the domain of $f(x) = \frac{1}{(x-2)^{1/4}} + (5-2x)^{1/2}$. > (Hint) Find the domains of $\frac{1}{(x-2)^{1/4}}$ and $(5-2x)^{1/2}$ separately, first. The domain of this function $f(x)$ will be the set of numbers these two domains have in common. 7. Verify that algebraic expression $$\frac{\sqrt{x^2-5\ }-2}{x-3}$$ is equal to the following algebraic expression (for any $x\ne 3$ and $x\ge\sqrt{5}$), $$\frac{x+3}{\sqrt{x^2-5\ }+2}.$$ > (Hint) There are a few ways to do this verification, but try to do it by multiplying the top and bottom of the original expression by the [*conjugate*](https://mathbitsnotebook.com/Algebra2/Radicals/RDMultDivide.html#:~:text=A%20conjugate%20is%20a%20binomial,is%20called%20a%20complex%20conjugate.) of $\sqrt{x^2-5\ } -2$. This would allow you to figure out the 2^nd^ expression if you were just given the 1^st^. > The reverse process is called "Rationalizing the expression." Review [video](https://youtu.be/AFY6ANog9rI). 8. Define $f(x) = \frac{\sqrt{x^2-5\ }-2}{x-3}$ and $g(x)= \frac{x+3}{\sqrt{x^2-5\ }+2}$. Find the domain of each of the functions $f$ and $g$. Do they have different domains? ::: spoiler They do not have the same domain. The domain of $f$ is $(-\infty,-\sqrt{5}]\cup[\sqrt{5}, 3)\cup(3,\infty)$. The domain of $g$ is $(-\infty,\sqrt{5}]\cup[\sqrt{5},\infty)$. ::: ### Asymptotes In general, graphs of algebraic functions can be complicated to determine. These functions will, sometimes, have vertical or horizontal asymptotes, and sometimes both. As we will discuss, saying that a graph has a particular asymptote is **the same** as saying that the function has a particular *limit*. To see more examples worked out, you might find some of the videos in [this playlist](https://www.youtube.com/playlist?list=PL3j1ntBPCU_r6fSJ7PQJXrze85bXa8oE3) helpful (for example, "Simplifying Rational Expressions" and "How to Simplify Radicals").