This week in Pre-Calculus 11 we continued to learn more about rational expressions. One new thing that I learned this week was how to add and subtract rational expressions.

Recall:

• Lowest Common Multiple (LCM):  the smallest number that two or more numbers divide evenly into. ex. The LCM of 2 and 3 is 6.
• What you do to the numerator you need to do to the denominator (what you do to the top you need to do to the bottom).
• In order to add and subtract fractions they need to have a common denominator.

Adding and Subtracting rational expressions:

• When adding and subtracting rational expressions it is important that they have a common denominator. This can be achieved by finding the LCM between the denominators of the fractions.

Example 1:

• First, you need to see if you can factor anything in the expression. In this case you would leave it as is since it can not be factored any further.
• Next, you look at the denominator and check if they are the same, if not you need to make it so that they have a common denominator. In this case since one denominator is 3x and the other is 5x, we know that the x’s are common but not the coefficient in front, we can fix this by multiplying each expression by the opposite coefficient. You would multiply the expression with 3x as the denominator by 5 and the one with 5x in the denominator by 3.
• After multiplying across the numerator and denominator, both expressions will have a common denominator. Due to this, you can write you expression as one big fraction since the denominators are now the same. You do this by combining the numerator of each expression (writing it as one continuous expression).
• Next you simplify the numerator if possible. In this case you simplify it by adding the two numbers in the numerator together.
• After you have completely simplified the expression, you need to make sure that you find the non-permissible values since the denominators contained variables. To do this you take a look at the factored form of the expression. In the factored form, the denominators were 3x and 5x, since this is the case we know that x≠0, otherwise the expression would be undefined since the denominator can never be equal to zero.

Example 2:

• First, you need to factor the expression if possible. If not then you would leave the expression as is.
• Next, you need to find the common denominator between the two expressions. In this case one expressions has a denominator of (x+2)(x-2) and the other has a denominator of (x+2), since both have x+2 in common, we know that the expression with the denominator of only (x+2) will need to be multiplied by (x-2) to make the denominators of the two expressions the same.
• Next, since we have multiplied to make the denominators the same, we can re-write the expression as one big fraction, by combining the numerators. Note, since this involves subtraction, the numerator of the expression that is being subtracted needs to have brackets around it. This is because of the negative in front that needs to be distributed in.
• After distributing the negative, simplify the expression if possible.
• After completely simplifying the expression, you need to find the non-permissible values, which you can find by taking a look at the factored form of the expression and the denominators of the factored expressions. In this case x≠ -2,2.

This week in Pre-calculus 11 we started a new unit that is all about rational expressions. At first we learned about equivalent rational expressions and then we learned how to multiply and divide them. What stuck with me was how to multiply and divide the rational expressions.

What to know:

• Rational Expression: The quotient of two polynomials ex.
• Non-Permissible Values: Similar to restrictions, Permissible values tell you what numbers x can not be equal to. These are only written if there are variables in the denominator, this is because we need to make sure that the denominator does not equal to zero since zero can’t be the denominator of a fraction. The non-permissible values are found using the zero product law. ex. x+8 = 0, x=-8 (when x = -8, the equation is equal to zero).

Example 1:

• First what you need to do is factor each fraction so that it is easier for you to simplify the expression.
• Next, simplify the expression. First what you want to do is need to check to see if there is anything in the numerator and denominator that matches (any pairs?), if so then you would cross them off since they cancel each other out, then once you have found all of the pairs you would multiply across both the numerator and denominator.
• After finding all the pairs in the question, you would re-write the fraction and see if their is anything that can be further simplified. Note: ex. expressions similar to this wouldn’t be simplified any further; this is because though it may be tempting to get rid of both 3’s, the x in the denominator is connected to the 3 by the +, this means that you would need to have something in common with both the x and the 3 to simplify it further.
• Next, since there are variables on the bottoms of the fractions, you find the non-permissible values. To do this we take a look at the original question (or its factored form) and use the product law and make the parts that involve variables equal to zero so that we can isolate and find x, these values will tell you what x can not be equal to.

Example 2:

 

• First, since this is a fraction being divided by a fraction we need to make it easier for us to solve. To do this you need to turn it into a multiplication problem, which is done by reciprocating the second fraction in the question.
• After reciprocating the second fraction, we can treat this like a multiplication question; so therefore you would factor the question, look to see if the numerator and denominator have anything in common (pairs) which you would then cross out since they cancel each other out, and then multiply across both the numerator and denominator.
• Next you would re-write the simplified expression and see if it can be further simplified.
• Now, if there are variables that are located in the denominator you need to find the permissible values. One way that dividing rational expressions differs from multiplying them is the non-permissible values. This is because when writing the non-permissible you need to take into account that both halves of the fraction that was reciprocated were the denominator at one point, therefore, you need to find the non-permissible values of both the numerator and denominator of the fraction that you reciprocated as well as the denominator of the fraction that you didn’t reciprocate.

This week in Pre-calculus 11 we learned more about absolute values and reciprocal functions. One thing that I learned that really stuck with me was when we learned about reciprocal functions and how to graph them.

Things to know:

• A positive reciprocates to a positive (+ –> +)
• A negative reciprocates to a negative (- –> -)
• When a big number is reciprocated it turns into a small number and vise versa.
• When both 1 and -1 are reciprocated they stay the same (called invariant points)
• When 0 is reciprocated it becomes undefined (you can’t have a 0 on the bottom of a fraction)
• Asymptote – horizontal/vertical line that separates the graph. The graph will approach both lines but never touch them. (the vertical asymptote is located halfway in-between the two invariant points/Horizontal will always be y=0).
• Hyperbola – two curves similar to each other, indicated by the similar points. Opposite of each other. Curved lines on the graph.
• The two hyperbolas will never touch due to the vertical asymptote

Example:

   

   

• First, we graph the linear function as we normally would if it wasn’t reciprocated. To do this we would write down everything we know about the original function and plot it on the graph. Since the original function is 2x+6, we know that the function will be positive and have a slope of 2, we also know that the y-intercept will be 6.
• Next, you circle your invariant points which will be when y is at both 1 (_,1) and at -1 (_,-1). To do this go to where 1 is on the y-axis and then slide your finger vertically along the graph until you touch the linear function. You would do the same thing for -1.
• Next you draw in the asymptotes. Since the horizontal asymptote will always be y=0, we draw a horizontal broken line along that part of the graph. Also, since the vertical line is half way in between the two invariant points, we know it will be when x= -3. Another way to find what the vertical asymptote is without looking at a graph is by making the original linear function equal to zero and solving for x.
• After drawing in the asymptotes, you can now draw the hyperbolas. To do this you would take of the points along the linear function, reciprocate it and plot its reciprocal. You would then draw the hyperbola through the invariant and the new reciprocated points.

Note: when writing the domain or range x and y can be any real number except they can’t be equal to the corresponding asymptote.

This week in Pre-calculus 11 we had our chapter 5 unit test and started unit 8. Something that I learned this week was how to graph the absolute value of both linear and quadratic functions, as well as how to write them in piecewise notation.

Recall:

• When finding the absolute value of something it is asking how far away a number is from zero. The absolute value will always be a positive number.

Need to know:

• The critical point (also called the point of inflection) is the point along the x-axis where the graph of the function changes direction.
• Piecewise notation, the first part of piecewise notation is used to state which parts of the original line are positive and didn’t need to be reflected, uses signs (≤ or ≥). The second part is used to describe the part of the function that had to be reflected upwards, uses signs (< or >).

Example 1:

• First, you graph the function as if there were no absolute value symbols around it. When graphing it normally we know that the y-intercept is 4 and that the linear function has a slope of 2.
• Next we graph the absolute value of the function. Since we already know that absolute values will always be positive we take the part of the function that runs past the x-axis and reflect it upwards. Reflecting it upwards will put it above the x-axis, making it positive (it’s the same as multiplying the negative points by -1). The part of the line that is already in the positive stays the same.
• After graphing the absolute value of the function you need to write the piecewise notation. first we write a notation for the portion of the function that stays the same. Since the right side of the graph doesn’t change we would write the notation as, 2x+4, x ≥ -2 (-2 is the critical point and doesn’t change). For the second part we write about the part of the graph that was reflected, when reflecting the values, it is the same as multiplying by -1. we would write the notation as -(2x+4), x < -2 (we only use less than because the critical point never changed).

Example 2:

•  Graph the function as if there were no absolute value symbols around it. When graphing it normally we know that the vertex is (3,-2) and that the quadratic function has a scale of 2-6-10.
• Next we graph the absolute value of the function. To do this we take the part of the parabola that is in the negative and reflect it upwards so that it is positive. To do this all you need to do is change the sign of the y in the vertex to a positive –> (3,2). The part of the function that is in the positive stays that same.
• After graphing the absolute value of the function you need to write the piecewise notation. First we write the notation about the part of the function that remained the same, we know that , x ≤ 2, x ≥ 4 (we write about the sides of the function since they didn’t change). We then write about the part of the function that was reflected upwards (multiplied by -1). We know that , 2 < x < 4 (in between those two points is where the function changes).