The slope-intercept equation (y = mx + b) is a linear equation that gives both the slope of a line and where the line crosses the y-axis. It has many applications outside of mathematics, from construction and describing the grade of terrain to describing the results of scientific experiments if there is a linear relationship between the independent and dependent variable.

The slope of a linear equation is the relationship between the change in the x variable and the change in the y variable. It is a ratio, defined as the change in y (the rise) over the change in x (the run). In symbol form, the slope of a line containing the points (x_{1, }y_{1}) and (x_{2, }y_{2}) equals the ratio of (y_{2}-y_{1})/(x_{2}-x_{1}). When solving a linear equation, the y coordinates must be subtracted in the same order as the x coordinates in order to solve the equation, and describe the line correctly. Suppose a line contains the points (1, 2) and (4, 4). The slope will equal the ratio of (4-2)/ (4-1) or 2/3.

Figure 1: The slope of a line is a ratio of the change in y (the rise) over the change in x (the run).

The slope of a line also tells how it slants. Suppose that m is positive, such as in the previous example of 2/3. The line slants upwards from left to right. If m is negative, the line slants downward from left to right. The larger the slant, the greater the slope. If the line is parallel to the x-axis, the slope is 0. A line that is parallel to the y axis, however, has an undefined slope, as it is like dividing by zero.

Figure 2: If there is no change in y, the slope of the line is 0, and the line is parallel to the x axis.

The y-intercept is the point where the line crosses the y-axis. Recall in a graph of Cartesian coordinates there is an x-axis (the horizontal one) and a y-axis (the vertical one). In a linear equation, the y-intercept is a single point, a constant.

Figure 3: The y intercept is a single point, a constant

Suppose that a hill has a grade of 10%. That is the rise over the run, or for every horizontal distance of 100 feet the road rises 10 feet. Snoqualmie Pass is the largest pass in Washington State that is kept open year-round, and a major east-west route over I-90. A route from Seattle to the summit of Snoqualmie pass goes from an elevation of 520 feet to elevation of 3022 feet, a rise of approximately 2502 feet. Although the distance between Seattle and Snoqualmie pass is about 50 miles, the grade varies and frequently exceeds the recommended average grade of 6% for interstate highways. This makes it more difficult for long-distance truckers carrying heavy loads between the eastern and western parts of the state, so that highway engineers are constantly modifying the highway to improve conditions.

Figure 4: Snoqualmie Pass is a major route over I-90, and is kept open year-round as road conditions allow.

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]]>Suppose we have an integral which appears unsolvable by ordinary means, such as

∫xsin(x)dx In many cases, we can do what is called integration by parts, which is where we split the equation we are taking the integral of into two parts, with the goal of simplifying the equation such that we can reduce it down to an equation we know how to solve.

Figure 1: General formula for Integration by Parts

In essence, what this means is that we are going to attempt to split up the equation in a way such that we can eliminate one of the equations inside the integral, allowing us to end up with a form that we know how to solve.

∫xsin(x)dx

Let u(x)=x and v’(x)=sin(x)

Then we need to find u’(x) and v(x), so u’(x) = d/dx(x) = 1

Next, we need to find v(x), which we do by doing ∫v’(x)dx, or in this case, ∫sin(x)dx = -cos(x)

Now that we have u(x), v(x), u’(x), and v’(x), we can plug everything in, resulting in the following:

∫xsin(x)dx = x*(-cos(x)) – ∫1*(-cos(x))dx

= -xcos(x)+∫cos(x)dx

= -xcos(x)+sin(x)+C

In general, we will want to treat the polynomial as u(x) and the other team as v(x) where possible. In addition, Integration by Parts can be done multiple times. Suppose we have the equation

∫x^{3}e^{x}dx

Then we will let u(x) = x^{3} and v’(x) = e^{x}, then u’(x) = 3x^{2} and v(x)=e^{x}, and we find that

∫x^{3}e^{x}dx = x^{3}e^{x} – ∫3x^{2}e^{x}dx

Then we can do the same procedure with the slightly easier integral ∫3x^{2}e^{x}dx

Let u(x) = 3x^{2} and v’(x) = e^{x}, then u’(x) = 6x and v(x) = e^{x}

Then ∫x^{3}e^{x}dx = x^{3}e^{x} – ∫3x^{2}e^{x}dx = x^{3}e^{x} – (3x^{2}e^{x} – ∫6xe^{x}dx)

And then with ∫6xe^{x}dx, we split it one last time into u(x) = 6x, v’(x) = e^{x}, and therefore u’(x) = 6 and v(x) = e^{x}

Then we can substitute in one last time in order to get ∫6xe^{x}dx

∫x^{3}e^{x}dx = x^{3}e^{x} – ∫3x^{2}e^{x}dx = x^{3}e^{x} – (3x^{2}e^{x} – (6xe^{x} – ∫6e^{x}dx)) =

x^{3}e^{x} – 3x^{2}e^{x} + 6xe^{x} – 6e^{x} + C. Whew!

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]]>Occasionally, you will have what are called composite functions; that is, functions that are composed of multiple functions and thus cannot be differentiated easily. In reality, although these may look tricky, they are actually fairly straightforward.

Suppose we have an equation written as f(g(x)). Then

d/dx f(g(x)) = f’(g(x))g’(x)

If the equation is more complicated, such as

d/dx(f(g(h(x))),

then it can be broken thought of as doing the chain rule multiple times.

Think of it instead as f(j(x)) where j(x) = g(h(x)), and as we know the derivative of f(j(x)) is f’(j(x))j’(x). So since j(x)=g(h(x)), we can rewrite this as f’(g(h(x)))*(g(h(x)))’, or f’(g(h(x)))g’(h(x))h’(x).

It can be thought of as a chain, in that first we take the derivative of the outer equation and then move steadily inward.

d/dx(sin(2x))

Then f(y)=sin(y) and g(x)=2x, so sin(2x) can be written as f(g(x)).

Then as explained before, d/dx(f(g(x)) = f’(g(x))g’(x)

=(sin(g(x)))’(2x)’

=cos(g(x))*2

=2cos(2x)

d/dx(3(cos(2x^{3}))^{4})

=3d/dx(cos(2x^{3}))^{4})

Let f(z)=z^{4}, g(y)=cos(y), and h(x)=2x^{3}

Then f’(z)=4z^{3}

g’(y)=-sin(y)

h’(x)=6x^{2}

And since the equation can be written as f(g(h(x))), then

3*d/dx(f(g(h(x))))=3*f’(g(h(x)))*g’(h(x))*h’(x)

Or, by substituting the values in,

3*4(cos(2x^{3}))^{3}(-sin(2x^{3})*6x^{2}

And, by combining like terms and

=-72x^{2}sin(2x^{3})(cos(2x^{3}))^{3}

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]]>Sometimes, a derivative is done using the chain rule, and it leaves with an equation that, at first glance, can look intractable when we are attempting to integrate it. In those cases, although the problem may look difficult or often will look like something that can only be solved through the use of Integration by Parts, there often exists a much simpler solution: that of U-Substitution.

Suppose we have an equation in the form ∫u’(x)f(u(x))dx. Then we can do a substitution in order to make the equation easier to integrate in a way that we will see shortly.

Although the way it was phrased above may make it sound intimidating, in reality the process is very straightforward.

Suppose we wish to find ∫x^{3}sin(x^{4})dx Although this may look tricky, once we notice that the derivative of x^{4} includes x^{3}, it becomes much more straightforward.

We let u=x^{4}

Then in that case, du/dx=4x^{3}, or du/4 = x^{3}dx

If we move things around in the initial integral, we can note that

∫x^{3}sin(x^{4})dxdx = ∫sin(x^{4})x^{3}dx

Now, since x^{4} = u, and x^{3}dx = du/4, we can substitute in for the x as such

∫sin(u)du/4 = 1/4 ∫sin(u)du

Which is ¼(-cos(u))+C = -¼cos(u)+C

And now we need to substitute back for u, so

-¼cos(u)+C = -¼cos(x^{4})+C

If all of the xs have not been eliminated, then we cannot go through with the substitution. A good choice of u will leave no x or dx in the equation at all, only u and du.

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]]>When the coefficient of the squared term is greater than 1, an extra step is added to factoring the trinomial. It can be factored by either using FOIL or a modified form of factoring by grouping.

When the squared term of a quadratic trinomial is x^{2} with a coefficient of 1, it always factors as (x + y)(x + z). If the coefficient is greater than 1, the factors of the squared term must be considered. Suppose the squared term is 2x^{2}. Its factors are 2x and x. If the squared term were 4x^{2}, it would have even more possibilities: x and 4x, 2x and 2x.

Figure 1: Factoring the squared term when the coefficient is greater than 1.

Factoring is the reverse of multiplying. The expression (2x + 3)(x + 5) is equal to 2x^{2} + 13x + 15. The product of the first terms is equal to 2x^{2}. The middle term is derived from the sum of the products of the outer terms, 10x, and the inner terms, 3x. The product of the last terms is equal to 15. Using this logic, the factoring for 2x^{2} + 7x + 6 can be derived similarly. The squared term can be factored as 2x times x. The constant has more possibilities, 1 and 6, 6 and 1, 2 and 3, or 3 and 2. For example, (2x + 1) and (x + 6) can be paired, or (2x + 6) and (x +1). Similarly, (2x + 3) and (x + 2) can be paired, or (2x + 2) and (x + 3). Suppose (2x + 1) and (x + 6) are paired. In that case, x + 12x equals 13x, which is too high. If (2x + 6) and (x + 1) are paired, then 2x +6x equal 8x, which is also too high. Suppose (2x + 2) and (x + 3) are paired. Then 6x and 2x equal 8x, which is also too high. That leaves (2x +3) and (x + 2). In that case 4x and 3x equal 7x, which is the correct alternative.

Figure 2: Choosing factors using FOIL.

When factoring by grouping, rewrite the trinomial with 4 terms rather than 3, as 2x^{2} + 3x + 10x + 15). The first group can be factored as x (2x + 3) and the second group as 5(2x + 3). Using the distributive property, the factors are (x + 5)(2x + 3), which is equivalent to (2x + 3)(x + 5). Similarly, the trinomial 2x^{2} + 7x + 6 can be rewritten as 2x^{2} + 3x + 4x + 6. The first group can be rewritten as x (2x + 3) and the second group can be rewritten as 2(2x + 3). Using the distributive property, the factors are (x + 2)(2x + 3).

Figure 3: Choosing factors by grouping.

Suppose the polynomial is 2x^{2} + 3x + 7. It is in the form of ax^{2} + bx + c, where a equals 2, b 3, and c 7. In order to choose factors by grouping, a∙ c should equal b. In this case a is 2 and c is 7, and some of the factors of 14 are 1 and 14 and 7 and 2. The sum of 1 and 14 is 15 and the sum of 7 and 2 are 9, neither of which is equal to 3. That polynomial is a prime polynomial.

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]]>Trinomials are a special type of polynomials that have three terms, usually a squared term, a coefficient times the term that is to the first power, and a constant. In this review, the squared term has a numerical coefficient of 1.

The type of trinomial when the squared term has the coefficient of 1 has the form x^{2} + bx + c, where b, c, and x are real numbers. In the trinomial x^{2} + 7x + 12, x^{2} has the coefficient of 1, because 1x^{2} is the same thing as x^{2} using the multiplicative identity of 1. The number 7 is the coefficient of the second term x, and 12 is the constant. In the trinomial x^{2} – 5x + 6, the coefficient of the second term is -5 and the constant is 6. In the trinomial x^{2} + 2x – 24, the coefficient of the second term is 2 and the constant is -24.

Figure 1: The squared term has the coefficient of 1, the second term has the coefficient of 7 and the constant is 12.

A trinomial is factored when the factors (x + y)(x + z) are determined. When the factors are multiplied using the FOIL method, the result is the original trinomial. If the trinomial can be factored, the constant will equal the product of the two terms yz. The coefficient of the second term will equal the sum of the outer terms and the inner terms. Suppose the trinomial is x^{2} + 7x + 12. The number 12 has several factors, 1 and 12, 2 and 6, 3 and 4. In order to choose between them, the sum of the pairs of factors must equal the coefficient of the second term, or 7. The factors 1 and 12 add up to 13, so they won’t work. The factors 2 and 6 add up to 8, so they won’t work. The factors 3 and 4 add up to 7, so the factored form of x^{2} + 7x + 12 must be (x +3)(x + 4). Using the FOIL method, (x + 3)(x + 4) is x^{2} + 4x + 3x + 12 or x^{2} + 7x + 12.

If the constant is positive, the factors of the constant must have the same sign, either positive or negative. The way to determine the sign of the factors of the constant is to look at the sign of the second term. Suppose the trinomial is x^{2} + 7x + 12. Both factors must be positive, and both 3 and 4 are positive numbers that add up to 7. Suppose the trinomial is x^{2} – 5x + 6. In this case, the sign of the second term is negative (-5), so both factors must be negative. The negative factors of 6 are -1 and -6, and -2 and -3. In this case, -2x and -3x add up to -5x, so the factored form is (x – 2)(x – 3). Using the FOIL method, (x – 2)(x – 3) is x^{2} -2x – 3x + 6 or x^{2} – 5x + 6.

Figure 3: Multiplication of positive and negative numbers.

If the constant is negative, the factors of the constant must have different signs. Suppose the polynomial is x^{2} + 2x – 24. Some of the factors of -24 are 1 and -24, 2 and -12, and 6 and -4. In this case, 1 and -24 add up to -23, 2 and -12 add up to -10, and 6 and -4 add up to 2. The factored form is (x + 6)(x – 4). Using the FOIL method, (x + 6)(x – 4) is x^{2} + 6x – 4x – 24.

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]]>Factoring by grouping is the best method to use when some terms in a polynomial share one common factor and some other terms in the same polynomial share another common factor. The common factors can then be factored out using the distributive property.

The first step is to determine if all the terms in a polynomial have a common factor. If they do, then the greatest common factor (GCF) can be factored. If they do not, determine if there are two terms in the polynomial that have one common factor and two other terms that have another common factor. Suppose the four-term polynomial is ax + bx + ay + by. Using the commutative property, it can be rearranged, so that ax and ay have the common factor a, and bx and by have the common factor b.

Figure 1: Rearranging terms in a polynomial using the commutative property.

The first two terms, ax and ay, can be factored as a(x + y). The second two terms of the polynomial, bx and by, can be factored as b(x + y). The new polynomial is a(x + y) + b(x + y). Suppose the four-term polynomial is 5m + 2w + mw + 10. It can be rearranged so that 5m + 10 + mw + 2w, or 5(m + 2) + w (m + 2).

Figure 2: Factoring grouped terms

Suppose the polynomial were xy + 3x – 2y – 6. The first two terms can be factored as x(y + 3), and the second two terms can be -2(y + 3). When there is a negative term, the sign must be changed when it is multiplied, so that -2 times a positive 3 is -6. Similarly, suppose the polynomial were x^{2} – 3x + x -3. The first group, x^{2} – 3x, could be factored as x(x – 3). The second group, x -3, actually has one factor in common. The identity factor, 1, can be used, so that the second group can be factored as 1(x – 3).

Figure 3: When factoring negative terms, remember that signs change when multiplying.

The polynomial in symbol form a(x + y) + b(x + y) can be factored using the distributive property as (a + b)(x +y). Similarly, the polynomial 5(m + 2) + w (m + 2) can be factored using the distributive property as (5 + w)(m + 2). The polynomial x (x -3 ) +1(x – 3) can be factored as (x + 1)(x – 3). The identity factor was used to illustrate the distributive property and factoring by grouping.

Figure 4: Use the distributive property after factoring by grouping.

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]]>Factoring a monomial from a polynomial is a process of finding the greatest common factor for the constants, the greatest common factor for the variable terms, and then using the distributive property to factor out the greatest common factor (GCF).

In symbol form, if a ∙ b = c, then both a and b are factors of c. Suppose the monomial is 10x^{4}. There are the factors of 10 and the factors of x^{4}. The factors of 10 are 1, 10, 5, and 2. The factors of x^{4} are x, x^{2}, x^{3}, and x^{4}. They can be combined as 10 ∙ x^{4}, 10x∙x^{3}, 10x^{2}∙ x^{2}, and so on.

Finding the greatest common factor of two constants is a matter of finding the factors of both constants and checking to see what factors are in common. For example, the number 30 can be factored as 2∙3∙5 and the number 42 can be factored as 2∙3∙7. Their GCF, or greatest common factor is 2∙3, or 6.

Figure 1: Finding the GCF of constants.

Finding the GCF of two or more variable terms is also a matter of finding the factors of the variable terms, and seeing what factors are in common. Suppose the terms are x^{5}y^{3}, x^{2}y^{2}, and x^{3}y^{2}. The GCF will be the highest power of the common variables, or x^{2}y^{2}. If the terms were xy^{3}, x^{4}, and xy^{3}, the GCF would be x. All three terms have x in common, because there is no y term in x^{4}, but the highest power is x^{1} or x.

Figure 2: Finding the GCF of terms.

The first step in the process of factoring a monomial from a polynomial is to find the GCF of the constants and the terms. Suppose the polynomial is 3x^{2} + 6x + 9. The GCF of the constants is 3, because 3∙1 is 3, 3∙2 is 6, and 3∙3 is 9. There is no x term in common, because 9 does not have a variable associated with it. The second step in the process is to use the distributive property to factor out the GCF from each of the terms of the polynomial, so that the result is 3(x^{2} + 2x + 3). For example, if the polynomial is 9x^{3} + 27x^{2} + 24x, the constants factor as 3∙3, 3∙9, and 3∙8, so the GCF is 3. There is also a common variable in each term of the polynomial. Therefore, the GCF is 3x, and the result is 3x (3x^{2} + 9x + 8).

Figure 3: Factoring a monomial from a polynomial.

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]]>Solving application problems is a process that includes understanding the problem, translating it into an equation, solving the equation, checking the answer, and answering the question. This process can be used to solve many different types of problems.

The first part of the process involves understanding what is being asked. This includes noticing any key words that refer to operations and any quantities that are in relationship to one another. It is important to have an idea of what sort of quantity will represent a solution. Suppose a problem asks how many points the first-place winner had. That would require just one answer. If the problem described that the first-place winner had 10 points more than the second-place winner, and 17 points more than the third-place winner, all three point values would be necessary to completely answer the question.

Figure 1: Setting up the application problem involves finding quantities in relationship to one another.

The second part of the problem involves putting the problem in symbol form. This includes choosing a letter to represent a variable, and writing down exactly what the variable stands for. Suppose that the point values of the three winners totaled 114. Let the point value of the first-place winner be x, the second-place winner x – 10, and the third-place winner x – 17. Then, x + x – 10 + x – 17 = 114.

Figure 2: Translating the problem into an equation involves putting the problem into symbol form.

If x + x – 10 + x – 17 = 114, then 3x = 114 + 27, or 3x = 141. So, x = 47, x – 10 = 37, and x -17 = 30. To check, 47 + 37 + 30 = 114. The answer makes sense, and fits the parameters of the problem. In this case, the original equation was x + x -10 + x – 17 = 114.

If the work is done to understand the problem before it is set up and solved, it is easier to answer the appropriate question. In this case, all three point values were necessary to completely answer the question. Suppose that more information were added to the problem. During the same contest last year, the point values of the first-place, second-place and third-place winners totaled 100, but the second place winner had 14 points less than the first-place winner, and the third-place winner had 18 points less than the first place winner. Which year did the third-place winner earn more points, and what was the difference? Last year, the equation was x + x – 14 + x – 18 = 100, so 3x = 132. The first-place winner earned 44 points, the second-place winner earned 44-14 or 30 points, and the third-place winner earned 28 points. However, the question asked has 2 parts. Last year, the third-place winner earned 28 points, and this year, the third-place winner earned 30 points. This year, the third-place winner earned more points, and the difference was a gain of 2 points.

Figure 3: Answering the question asked brings all the parts of the problem into balance.

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]]>The first step in solving real-world problems mathematically is to change the problem into an equation. Sometimes, it is a matter of deciding what operations need to be used. It is also a matter of seeing how quantities relate to one another.

Words and phrases that represent addition include such things as added to, increased from, more than, or the sum of. For example, 9 added to a number could be represented algebraically as 9 + x. Words and phrases that represent subtraction include, subtracted from, less than, decreased by, or the difference between. Statements such as multiplied by, the product of, twice or three times, or a certain percent indicate multiplication, and statements such as divided by or the quotient of can indicate division. Sometimes a statement can indicate more than one operation. For example, a statement such as 5 less than 3 times a number could be represented as 3x – 5.

Figure 1: Some words that suggest mathematical operations.

It is also important to determine the relationship between quantities. Suppose that the length of a rectangle is 5 inches longer than its width. The width of the rectangle could be represented as w, so the length is w + 5. When one quantity is given in terms of another, it is usually easier to let one variable stand for the base amount. Any letter other than x can be used, as long as what the variable stands for is clearly defined.

Suppose that one notebook costs $3.00. Then 5 notebooks would cost 5 ∙ 3 or $15.00. If one textbook costs x, then 5 copies of the same text would cost 5x. Similarly, the cost of x items at y dollars can be represented as xy. Sometimes a statement can include more than one operation. Suppose the cost to rent a truck included a daily fee of $40.00, as well as a mileage fee of 60 cents a mile. If the truck were being rented for 2 days, the total cost of the truck would be the daily fee of 40.00 ∙ 2 for the number of days, plus the mileage fee of 0.60x.

Figure 2: Using multiplication and addition to solve a problem.

The word is often means is equal to. Suppose in the truck rental example, the total cost of the truck rental is $260.00. Then $80.00 + 0.60x = 260.00. The equation could then be solved as .60x = 260 – 80, or .60x = 180, or x = 180/.60 = 300 miles.

Figure 3: An example of translating an application to an equation.

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