Learning objective
Rational functions can also be rewritten in various forms to expose different characteristics. This can include vertical asymptotes, horizontal asymptotes, or slant asymptotes. Polynomial long division can be used to rewrite a rational function, allowing for the identification of slant asymptotes. Because rational functions are the quotient of two polynomials, the same strategies we used to rewrite polynomials can be used to rewrite rationals.
Find the slant asymptote of h\left(x\right) = \dfrac{3x^3 - x^2 + 2x - 5}{x^2 - 1}.
By rewriting a rational function in different forms, we can identify various characteristics such as vertical and slant asymptotes.
In geometry, we found combinations by using the following formula:
This formula uses factorials.
Calculate the following values:
Compare your answers to the triangle below.
The triangle we will explore today was discovered by French mathematician Blaise Pascal in the 17th century. Although we call it Pascal's triangle, students from other areas of the world know it by a different name.
Back in those days, there was often no means of sharing something published in one country with other countries. Because of this, mathematicians all over the world often discovered concepts around the same time and independent of each other. There are other names for it in other countries, but we call it Pascal's triangle.
Although there are many interesting mathematical phenomenons found within the triangle, the one we will focus on today is it's relationship to combination. In general, we can find the values of {}_n C_r in row number n (starting at row 0) and the r value is the element in the row, (also starting at 0).
So, the value of {}_4C_2 is found in row 4 (remember to begin counting from 0) and is in position 2 (count from left to right starting from 0). We see {}_4C_2=6 and can confirm this using the combination formula:
\displaystyle {}_4C_2 | \displaystyle = | \displaystyle \dfrac{4!}{(4-2)!2!} |
\displaystyle = | \displaystyle \dfrac{4!}{2!2!} | |
\displaystyle = | \displaystyle \dfrac{4\cdot3\cdot2\cdot1}{2\cdot1\cdot2\cdot1} | |
\displaystyle = | \displaystyle 6 |
The notation \binom{n}{r} may also be used for combinations.
Find row 10 of Pascal's triangle.
Evaluate each expression using the given approach.
Evaluate \binom{6}{3} using Pascal's triangle.
Evaluate {}_{7}C_{4} using the combination formula.
We can find the value of {}_nC_r or \binom{n}{r} using the combination formula
or by using Pascal's triangle. In general, we can find the values of {}_n C_r in row number n (starting at row 0) and the r-value is the element in the row, (also starting at 0).
Evaluate the following binomials:
To expand binomials in the form (a+b)^n, we previously used the distributive property to multiply {(a+b)} to itself n times. However, there are patterns in the expansions, and these patterns can be summarized into the binomial theorem.
In general, the binomial theorem says:
The symbol \sum is captial sigma, and we call it a summation. It tells us to begin by substituting the lower limit (r=0 for the binomial theorem) into the expression on the right side. This corresponds to the first term in the expansion. Then, we substitute r=1 for the second term, r=2 for the third term, and continue this pattern until we substitute r=n, the upper limit, for the (n+1)th term. Because it is a sum, we add each of the terms together.
\begin{aligned}\left(a+b\right)^n&=\sum_{r=0}^{n}\binom{n}{r}\left(a\right)^{n-r}\left(b\right)^r\\&=\binom{n}{0}a^nb^0+\binom{n}{1}a^{n-1}b^1+\ldots +\binom{n}{n-1}a^1b^{n-1}+\binom{n}{n}a^0b^n\end{aligned}
For example: \begin{aligned}(a+b)^4&=\sum_{r=0}^4\binom{4}{r}\cdot a^{4-r}\cdot b^r\\&=\binom{4}{0}\cdot a^{4-0} \cdot b^0 +\binom{4}{1}\cdot a^{4-1} \cdot b^1+\binom{4}{2}\cdot a^{4-2} \cdot b^2+\binom{4}{3}\cdot a^{4-3} \cdot b^3 +\binom{4}{4}\cdot a^{4-4} \cdot b^4\\&=a^4+4a^3b+6a^2b^2+4ab^3+b^4\end{aligned}
Looking at the exponents on the variables in an expansion, the exponents of the first term are integers that descend beginning from n and ending at 0. The exponents of the second term are integers that ascend beginning from 0 and ending at n.
Because the coefficients in the binomial theorem are combinations, the elements of Pascal's triangle can be used to evaluate the coefficients as row n will given the coefficients of the terms of \left(a+b\right)^n in order.
We can use the binomial theorem to find a specific term in the expansion without having to fully expand the binomial. Each term in the expansion of (a+b)^n is of the form: \binom{n}{r}\cdot a^{n-r}\cdot b^r where r corresponds to the (r+1)th term since we begin couting the terms from 0.
So if we need to find the 10th term, we let r=9.
When fully expanded, each expansion will have (n+1) terms because we begin counting the terms from 0. So if we need to find the second to last term and n=12, there will be 13 total terms and the second to last term will be the 12th term. In this case, r=11.
Use the binomial theorem to expand the following expressions.
(5x-7y)^2
\left(p+4\right)^{5}
Find the middle term in the expansion \left( 2x^{3} - 3y^{2}\right)^{8}.
Find the term in the expansion of \left(2a-\dfrac{b}{3}\right)^9 that contains a^6.
The binomial theorem says:
Each individual term is of the form\binom{n}{r}(a)^{n-r} (b)^rand we can use this expression to find the (r+1)th term of the expansion.