Integration problems involving simple substitutions

It might be helpful to see the basic integration formulas and how simple substitution is done, as discussed here, and here, since the following simple substitution problems are based on them.

Substitution is a very important concept in integration. If you want to learn integration, you need to learn substitution. Basically substitution is just spotting an appropriate expression in the given integral and then putting another variable in place of that expression. It involves a few rules, which will be clear from the following examples:

Example 1

`\int sin(x)cos(x) dx`
First, let us determine whether the above problem can be solved by using any of the basic integration formulas. There are integration formula for sin(x), and one for cos(x), but none for sin(x)cos(x). We are already familiar with basic substitution in integrals.

Now what we have to do is that we have to look for an expression in this integral, whose differential coefficient (or, simply, derivative) will give us the rest of the integral. Let us consider sin(x). The derivative of sin(x) is cos(x). Thus if we substitute
`t = sin(x)`
then
`dt = cos(x) dx`
So we can write the integral as,
`\int sin(x) cos(x) dx = \int t dt`
(if you are unable to understand this step then have a look at this post which discusses it in detail)
Now can you integrate `\int t dt` ? If you know the power rule, you can do this quite easily:
`\int t dt = t^2 / 2 + C`
The last step of any substitution, except possible in definite integrals, is to replace the original substitution. Otherwise, your answer is incorrect. Therefore,
`\int sin(x) cos(x) dx = sin^2(x) / 2 + C`     ... (Answer)

Example 2

 `\int sin(2x) dx`
This is quite easy. We know the formula `\int sin(x) dx = -cos(x) + C`, so we directly get `\int sin(2x) dx = -cos(2x) + C`, don't we? Actually no. The integration formula can not be directly applied if you have any expression in place of the single variable in the formula. Here we have `2x` in place of `x`. So the formula can't be used directly.

So we simply substitute
`t = 2x`.
Then dx is obtained by differentiating both sides of the equation with respect to x:
`dt = 2 dx`
`1/2 dt = dx`
So we replace 2x with 't' and `dx` with `1/2 dt` to get
`\int sin(t) 1/2 dt`
Numbers can be factored out of the integral (that is, if they are in multiplication),
`= 1/2 \int sin(t) dt`
Then applying the formula `\int sin(x) dx = -cos(x) + C`, we get
`= 1/2 (-cos(t)) + C`
Now replace t with 2x and simplfy to get:
`= -cos(2x)/2 + C`    ... (Answer)

Basic integration formulas

The following are some basic integration formulas. (You need to learn them in order to be able to do integration)

Trigonometric formulas

`\int sin(x) dx = -cos(x) + C`
`\int cos(x) dx = sin(x) + C`
`\int sec^2(x) dx = tan(x) + C`
`\int csc^2(x) dx = -cot(x) + C`
`\int sec(x)tan(x) dx = sec(x) + C`
`\int csc(x)cot(x) dx = -csc(x) + C`
Notice that the formulas for sec(x), cot(x), tan(x) and csc(x) are missing in the above list. This is because their formulas are more complex and are derived using substitutions and basic integration formulas. So they are discussed later.

Exponential function

`\int e^x dx = e^x + C`
`\int a^x dx = a^x /ln(a) + C` (where 'a' is a positive number greater than zero)
It may also be beneficial to again state the two formulas we have learnt till now:
`\int x^n dx = x^(n + 1) /(n + 1) + C`
`\int 1/x dx = ln(x) + C`

Simple substitution in integrals

All integrals can not be evaluated by directly using the integration formulae. Substitution is an important part of integration. It helps you to convert an integral into a 'nicer' form which will help you apply one or the other integration formulas, directly to it.

Substitution in integrals does not follow a rigid rule. Only by looking at the integral properly can you get to know what to substitute for what.

First, let us understand what substitution means. In integration, substitution means putting a variable in place of an expression. For example, in the following integral
`\int (2x + 1)^10 dx`
if we put `t` in place of `2x + 1` then it becomes quite simple to evaluate the integral because we get `t^10`, which can be evaluated using the power rule of integration.

But it is a rule in integration that the variable in 'dx' part of the integral should be the same as the variable of integration. Since we have changed the variable of integration from `x` to `t` by substituting `t = 2x + 1`, therefore we have to change `dx` to `dt` by some method.

The method to change `dx` to `dt` is to find the differential coefficient of the substitution. That is, we will differentiate both sides of the equation `t = 2x + 1` to get:
`d/dx t = d/dx (2x + 1)`
`dt/dx = 2`
Then solve for `dx`,
`dx =  1/2 dt`
So now we know that we can write `1/2 dt` in place of `dx`. The above integral, then, can be written as
`\int t^10 1/2 dt`
Factoring 1/2 out of the integral and applying the power rule of integration,
`1/2 \int t^10 dt`  ... (factoring out 1/2)
`1/2 t^11 /11 + C`   ... (applying power rule)
`1/22 t^11 + C` 
Now comes the most important step: Replacing `t` with the original substitution `2x + 1`.
`1/22 (2x +1)^11 + C`     ... (Answer)
Therefore we can write,
`\int (2x + 1)^10 dx = 1/22 (2x +1)^11 + C`
Remember that substitution in integrals does not follow a specific rule as to what you have to substitute for. Sometimes it may be apparent, and sometimes it may take time to spot the substitution, especially in bigger integrals. As you progress from simpler to more complex substitutions, gradually you will be able to do substitutions quite quickly.

For integrals in which you have to substitute linear expressions (all of the above examples are of that kind), you can remember a shortcut method to solve them quickly: The linear expression being substituted may have a coefficient with 'x'. For example coefficient of x is 2 in the expression 2x + 3. So what you have to do is just put `t` in place of the linear expression, write dt in place of dx, and integrate it. The important part is that you don't have to find 'dx', which saves you time. Then after finding the integral and replacing `t` with the original substitution, divide the whole integral by the coefficient of x you noted above.

Next Post: Basic integration formulas

Integrating x^-1 or 1/x

You may be tempted to apply the power rule of integration to `x^-1`, but you can't.

This can be understood by trying to apply the power rule to `x^-1` and seeing what you get. If you apply the power rule to `x^-1`, on integrating you get division by zero:
`\int x^-1 dx = x^(-1 + 1) / (-1 + 1) = x^0 / 0 = 1/0`
Division by zero does not give you a well defined answer, so this method is not appropriate for integrating `1/x`. Instead there is a formula, as follows:
`\int 1/x dx = ln|x| + C`
... where `ln|x|` is the natural logarithmic function. It is same as `log_e|x|`. Note that `ln|x|` has absolute value bars around 'x' in the above formula. Examples usages of this formula are:

1. `\int 1/(2x) dx = ln|2x| /2 + C`
2. `\int 1/(3x + 7) dx = ln|3x + 7| /3 + C`
3. `\int 1/(x - 1) dx = ln|x - 1| + C`

Note: Simple substitution is used in the above example, which is discussed in the next post.

Next Post: Simple substitution in integrals

Integration - Power Rule

Just as you have a power rule in differentiation, you have one in integration. Further, since integration is exactly the opposite of differentiation, the method you use while integrating simple powers of 'x' is exactly the opposite while differentiating them.

Recall that while differentiating simple powers of `x`, like `x^2`, you multiply with the power itself (which is 2 here) and then decrease it by 1. Thus `d/dx x^2 = 2x`. In integration, you increase the power by 1 and divide by the new number. So `\int x^2 = x^3/3 + C`.

The power rule of integration can be understood in two steps:
  1. Increase the power (exponent) by 1
  2. Divide by the new power (exponent)
It can be stated as:
`\int x^n dx= x^(n+1)/(n+1) + C`
... where 'n' is any real number except -1 (that is, it can be any negative or positive number except -1 and even a fraction, an irrational number or a decimal)

Note: The power rule can not be used to integrate`x^-1`. This is because when you increase -1 by 1, you get 0. Thus you get `\int x^-1 dx = x^0 / 0` in which, as you can see, there is division by zero. Division by zero does not give a well defined answer. This rule is used instead.

Examples

1. `\int x^100 dx`
`= x^(100 + 1)/(100 + 1) + C `
`= (x^101)/101 + C`
2. `\int x^(-10) dx`
`= x^(-10 + 1) / (-10 + 1) + C`
`= x^-9 / -9 + C`
3. `\int 1/x^2 dx`
`= \int x^-2 dx`
`= x^(-2+1) / (-2 + 1) + C`
`= x^(-1) / (-1) + C`
` = -1/x + C`
4. `\int x^(2/3) dx`
`= x^(2/3 + 1) / (2/3 + 1) + C`
`= x^(5/3) / (5/3) + C`
`= 3/5 * x^(5/3) + C`
5. `\int sqrt(x) dx`
`= \int x^(1/2) dx`
`= x^(1/2 + 1) / (1/2 + 1) + C`
`= x^(3/2) / (3/2) + C`
`= 2/3*x^(3/2) + C`
6. `\int 1/sqrt(x) dx`
`= \int x^(-1/2) dx`
`= x^(-1/2 + 1) / (-1/2 + 1) + C`
`= x^(1/2) / (1/2) + C`
`= 2*sqrt(x) + C`
7. `\int x^log(2) dx`
`= \int x^(log(2) + 1) / (log(2) + 1) + C` ( log(2) is a number )
8. `\int x^sqrt(3) dx`
`= \int x^(sqrt(3) + 1) / (sqrt(3) + 1) + C`
Next Post: Integrating 1/x 

Indefinite Integrals

If you already know how to find the derivative of a function, understanding indefinite integrals becomes easier. An indefinite integral is defined as the parent function of a derivative. For example, if you have the function
`f(x) = 3x^2 + 4x + 5`
its derivative is
`f'(x) = 6x + 4`
Then we say that `f(x)` is the integral of `f'(x)`. Hence, integration is opposite of differentiation.

Now suppose we have the function
`f(x) = 3x^2 + 4x + 10`
its derivative is
`f'(x) = 6x + 4`
Notice that the derivative of `f(x) = 3x^2 + 4x + 10` is same as that of the derivative of `f(x) = 3x^2 + 4x + 5`. In other words, the integral of `f'(x) = 6x + 4` can be either `f(x) = 3x^2 + 4x + 10` or `f(x) = 3x^2 + 4x + 5`. This implies that one function can have more than one integrals.

In fact, one function can have an infinite number of integrals. Each of these integrals differs from the others only by a constant. As seen above, the two functions `f(x) = 3x^2 + 4x + 5` and `f(x) = 3x^2 + 4x + 10` differ only by a constant; the former has 5 and the latter has 10 instead.

The above fact is easily understood by the help of differentiation. We know that the derivative of a constant is zero. So in both the functions above, which differ only by a constant, the derivative of the constant does not matter since it is zero. Hence both have the same derivative. In opposite sense, both the functions are integrals of the same function, their derivative.

This is an important property of indefinite integrals. There can be an infinite number of indefinite integrals of a given function. Hence they are given the name 'indefinite'. This is also the reason that the proper way to represent an indefinite integral is to always put a variable '+ C' after the function in place of any constants present in the function. For example, as from the above examples, the integral of the function
`f'(x) = 6x + 4`
is the function
`f(x) = 3x^2 + 4x + C`
Notice that we put a '+ C' at the end instead of a + 5 or a + 10. This is the correct representation of an indefinite integral; it always has a trailing + C in it in place of any constants. This is a rule, which, if not followed, can get you incorrect answers.

The + C we put in the above integrals is called the constant of integration.

Now we should become familiar with some terms in integration:

  • Indefinite Integrals are also called antiderivatives because they are the result of a mathematical operation which is opposite that of finding the derivative of a function
  • Indefinite Integrals are also called primitive functions because in any process of differentiation, they are the original functions whose derivative is obtained.
  • The process of finding the integral of a function is called integration
  • Integrand  is the function being integrated. It is the expression between the `\int` and `dx` sign.

Representation of Integration:

The symbol for integration is `\int`. It is put before the function whose integral you have to find. Note that in integration, you need to put another symbol `dx` at the end of the integral which denotes the termination of the function being integrated. For example, integration of `x^2 + 4x + 1` is represented as 
`\int x^2 + 4x + 1 dx = 2x + 4 + c`
Again note the three remarkable things about the above equation:

  1. It starts with the integration symbol `\int`
  2. There is a `dx` denoting the end of the function which you are integrating
  3. There is a trailing '+ c' at the end of the evaluated indefinite integral
Also note that the function between the `\int` and `dx` sign is the integrand. These three rules are essential in writing any integration statement.

A Note on `dx`

The symbol `dx` is always written after the integrand. It represents a sort of a full stop. When a sentence is over you put a full stop after it as a punctuation mark. Similarly in integration you put a `dx` to represent that the integrand is only till here.

More importantly, this is not the only use of `dx`. It is mathematically very significant. The variable which comes after `d`, which, here, is `x`, is called the variable of integration. It is the variable with respect to which the integral is being evaluated. For example, the following integration statement:
`\int x^2 + 2x dx = 2x + 2 + c`
is read as "Integral of `x^2 + 2x` with respect to `x` is equal to ...". This is quite important. If suppose there is 'y' in place of 'x' in it, that is, 
`\int x^2 + 2x dy`
the above expression is a totally different from the previous one in which we had `dx`. Since here have a `dy`, it means the integration is being done with respect to 'y', and any other variable, like 'x', will be treated like a constant. That is, in the process of integrating the above expression, we will treat 'x' as a number not a variable because the expression is being integrated with respect to 'y'. Hence the integral will beome:
`\int x^2 + 2x dy = (x^2 + 2x)y + c`
Thus, the `dx` has a very important significance is specifying the variable of integration.


Why 'Indefinite Integral'? Why not just 'Integral'?

There are two types of integrals, definite and indefinite. Before learning definite integrals you should be acquainted with the indefinite ones. (definite integrals are nothing but a numerical extension of indefinite integrals). In this introductory post we are clearly specifying 'indefinite integral'. However, saying just 'integral' generally means the indefinite integral.

Rules and formulas to do integration 

Since the process of integration is exactly opposite to that of differentiation, most of the formulas for derivatives, if reversed, can give you formulas for integration. Some of them are:

Integrating simple powers of x

The first one is opposite to the power rule in differentiation. The power rule of differentiation says that `d/(dx) x^n = nx^(n-1)`. Its opposite in integration is:
`\int x^n dx = x^(n+1)/(n+1) + C`
(In differentiation you take the power, multiply it with x and then decrease the power by 1; In integration, you take the power, increase it by 1, then divide x by it as well.)

Exception: This rule can be used for all values of `n` except -1. If you want to integrate x^-1, you can't apply this rule. The rule discussed after this one will help you integrate that.

Notice the `+ C` at the end of the integral. It denotes a constant and is called the constant of integration. Any indefinite integral must always be written with a trailing `+ C` otherwise it is considered incomplete.

Example usages:
  1. Whole number powers:
    1. Simple without coefficients:
      1. `\int x^2 dx = x^(2 + 1)/(2 + 1) = x^3/3 + C`
      2. `\int x^3 dx = x^(3 + 1)/(3 + 1) = x^4/4 + C`
    2. With coefficients:
      1. `\int 3x^2 dx = 3x^(2 + 1)/(2 + 1) = 3x^3/3 = x^3 + C`
      2. `\int x^4/5 dx = x^(4 + 1)/(5*(4 + 1)) = x^5/25 + C`
      3. `\int -5x^2 dx = -5x^(2 + 1)/(2 + 1) = -5x^3/3 = -5x^3/3 + C`
  2. Negative whole number powers:
    1. Simple without coefficients:
      1. `\int x^-2 dx = x^(-2 + 1)/(-2 + 1) = x^-1/(-1) = -1/x + C`
      2. `\int x^-3 dx = x^(-3 + 1)/(-3 + 1) = x^-2/(-2) = -1/(2x^2) + C` 
    2. With coefficients:
      1. `\int 3x^-2 dx = 3x^(-2 + 1)/(-2 + 1) = 3x^-1/(-1) = -3/x + C`
      2. `\int -4x^-3 dx = -4x^(-3 + 1)/(-3 + 1) = -4x^-2/(-2) = 2/(x^2) + C`
      3.  `\int x^-4/5 dx = x^(-4 + 1)/(5*(-4 + 1)) = -x^-3/15 + C`
      4. `\int -(1/5)x^-4 dx = (-1/5)x^(-4 + 1)/(-4 + 1) = (1/15)x^3 + C`
  3. Fractional/real number powers:
    1. Simple without coefficients:
      1. `\int x^(1/2) dx = x^(1/2 + 1)/(1/2 + 1) = x^(3/2)/(3/2) = 2/3 x^(3/2) + C`
      2. `\int x^(4/3) dx = x^(4/3 + 1)/(4/3 + 1) = x^(7/3)/(7/3) = 3/7 x^(7/3) + C`
    2. With coefficients:
      1. `\int 2x^(1/2) dx = 2x^(1/2 + 1)/(1/2 + 1) = 2x^(3/2)/(3/2) = 4/3 x^(3/2) + C`
      2. `\int -1/2 x^(4/3) dx = -1/2 x^(4/3 + 1)/(4/3 + 1) = -1/2 x^(7/3)/(7/3) = -3/14 x^(7/3) + C`

Integrating `x^-1`

The rule for integrating `x^-1` is:
`\int x^-1 dx = log|x|`
Notice that there are absolute value bars in the logarithm. Further, the logarithm denoted has a base of 'e', that is, it is the natural logarithm. All integration rules are stated in terms of natural logarithms.

Example usage:
`\int x^-1 dx = log|x| + C`

Integrals of Exponential functions

As the derivative `e^x` is `e^x`, so the integral of `e^x` is `e^x`. Thus,
`\int e^x dx = e^x + C`
where `e` is a constant.

Further, as the derivative of `a^x` is `a^x log(a)`, so we can say that the integral of `a^x log(a)` is `a^x`. That is, `\int a^x log(a) dx = a^x/log(a)`. Since we want to get the integral of `a^x`, divide by log(a) on both sides of this equation to get `\int a^x dx = a^x/log(a) + C`. Thus the integration formula for the exponential function `a^x`, where a can be any number greater than zero, is
`\int a^x dx = a^x/log(a) + C`
Example usages:
  1. Exponential function with base 'e'
    1. `\int e^x dx = e^x + C`
    2. `\int `5e^x dx = 5e^x + C
    3. `\int -3e^x dx = -3e^x + C`
    4. `\int 1/3 e^x dx = 1/3 e^x + C`
  2. Exponential function with base 'a'
    1. `\int a^x dx = a^x/log(a)`
    2. `\int 2^x dx = 2^x/log(2)`
    3. `\int (1/2)^x dx = (1/2)^x/log(1/2)`
Note that the exponential function with any numbers as its base can only have positive numbers as the base.

Integrals of Trigonometric functions

Have a look at the differentiation formulas for trigonometric functions. If you take the opposite of each formula you get the integration formulas for trigonometric functions. For example, `d/(dx) sin(x) = cos(x)` means that `\int cos(x) dx = sin(x) + C`. Similarly for other trigonometric functions:

Derivative formulas for trigonometric functions
Integration formulas for trigonometric functions
`d/(dx) sin(x) = cos(x)`
`\int cos(x) dx = sin(x) + C`
`d/(dx) cos(x) = -sin(x)`
`\int sin(x) dx = -cos(x) + C`
`d/(dx) tan(x) = sec^2(x)`
`\int sec^2(x) dx = tan(x) + C`
`d/(dx) cot(x) = -cosec^2(x)`
`\int cosec^2(x) dx = -cot(x) + C`
`d/(dx) sec(x) = sec(x) tan(x)`
`\int sec(x) tan(x)(x) dx = sec(x) + C`
`d/(dx) cosec(x) = -cosec(x) cot(x)`
`\int cosec(x) cot(x) dx = -cosec(x) + C`
Example problems are discussed in a separate post.

Integration formulas from derivatives of inverse trigonometric functions

Let us take an example to understand this section. The derivative of `sin^-1(x)` is `1/sqrt(1 - x^2)`. Thus, the integral of `1/sqrt(1 - x^2)` is `sin^-1(x) + C`. Similarly some integration formulas are derived from differentiation formulas of inverse trigonometric functions. These are as follows:

Derivative formulas for inverse trigonometric functions
Integration formulas derived from them
`d/(dx) sin^-1(x) = 1/sqrt(1 - x^2)`
`\int 1/sqrt(1 - x^2) dx = sin^-1(x) + C`
`d/(dx) cos^-1(x) = -1/sqrt(1 - x^2)`
`\int -1/sqrt(1 - x^2) dx = cos^-1(x) + C`
`d/(dx) tan^-1(x) = 1/(1 + x^2)`
`\int 1/(1 + x^2) dx = tan^-1(x) + C`
`d/(dx) cot^-1(x) = -1/(1 + x^2)`
`\int -1/(1 + x^2) dx = cot^-1(x) + C`
`d/(dx) sec^-1(x) = 1/(x*sqrt(x^2 - 1))`
`\int 1/(x*sqrt(x^2 - 1)) dx = sec^-1(x) + C`
`d/(dx) cosec^-1(x) = -1/(x*sqrt(x^2 - 1))`
`\int -1/(x*sqrt(x^2 - 1)) dx = cosec^-1(x) + C`

Integration formulas for tan(x), cot(x), sec(x) and cosec(x)

If you look at all the differentiation formulas for trigonometric functions, there is no one formula in which any of these functions is the derivative itself. For example, the derivative of tan(x) is sec^2(x) but there is no trigonometric function whose derivative is tan(x). Thus we are unable to simply get the formula for integrating these functions. Their formulas are derived by means of the use of substitution, an advanced technique on integration. Thus these formulas are discussed in a separate post after substitution.

Advanced techniques of integration

Integration involves the use of advanced algebraic techniques in addition to the above formulas, unlike differentiation. These techniques are:
  1. Substitution
  2. The use of Partial Fractions
  3. Integration by Parts
No comments:

Solving and factoring quadratic equations

Solving Quadratic Inequalities

Quadratic inequalities are mathematical statements of comparison in which a quadratic expression is set greater than or equal to another quadratic or lower degree expression.

For example `3x^2 + 5x + 6 ≥ x + 5` is a quadratic inequality because the highest degree on any term is 2.

Solving quadratic inequalities:

Move all terms to the left side

`3x^2 + 5x + 6  - x - 5 ≥ 0`
`3x^2 + 4x + 1 ≥ 0`

Find the zeros of the quadratic polynomial on the left side

`3x^2 + 4x + 1 = 0`
`3x^2 + 3x + x + 1 = 0`
`3x(x + 1) + 1(x + 1) = 0`
`(x + 1)(3x + 1) = 0`
`x + 1 = 0, 3x + 1 = 0`
`x = -1, x = -1/3`
If the quadratic expression doesn't factor, then find its zeros by the quadratic formula. If the quadratic expression doesn't have any zeros then evaluate it at any value of 'x' to see whether it is greater than or lesser than zero.

Make a number line and divide it into segments by the zeros


Take values in between the zeros and before and after the zeros and plug them in the quadratic polynomial to evaluate it.

`x = -2`
`3x^2 + 4x + 1 = 3(-2)^2 + 4(-2) + 1 = 5`

`x = -1/2`
`3x^2 + 4x + 1 = 3(-1/2)^2 + 4(-1/2) + 1 = 3/4 - 2 + 1 = 3/4 - 1 = -1/4`

`x = 0`
`3x^2 + 4x + 1 = 3(0)^2 + 4(0) + 1 = 1`

Wherever the quadratic polynomial evaluates negative, it is lesser than zero and wherever it evaluates positive it is greater than zero. Notice that the quadratic inequality has a greater than zero sign. So the regions of the number line where the quadratic polynomial evaluates positive are the answers. These are
  • Before `-1: (-∞, -1)`
  • After `-1/3: (-1/3, ∞)`
Now since the inequality symbol also includes the equal sign as it is "greater than or equal to (≥)", so the zeros are also taken up. To include the zeros -1 and -1/3, put square brackets near them:

So the answer is `(-∞, -1] U [-1/3, ∞)`
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Graphing quadratics in Intercept Form - Solved Examples

Graph `y = (x - 1)(x - 4)`

Graph of `y = (x - 1)(x - 4)`
Comparing the equation with intercept form `y = a(x - p)(x - q)`, p = 1 and q = 4. So the x-intercepts are `(1, 0)` and `(4, 0)`.

Let the vertex be (h, k), then `h = (p + q)/2 = (1 + 4)/2 = 5/2`,

`k = (5/2 - 1)(5/2 - 4)`
`k = (3/2)(-3/2)`
`k = -9/4`

Therefore the vertex is `(5/2, -9/4)`. Now plot the vertex and x-intercepts and join them with a free hand curve

Graph `y = 3(x - 6)(x - 3)`


Graph of `y = 3(x - 6)(x - 3)`
Comparing the equation with intercept form `y = a(x - p)(x - q)`, p = 6 and q = 3. So the x-intercepts are `(6, 0)` and `(3, 0)`.

Let the vertex be (h, k), then `h = (p + q)/2 = (6 + 3)/2 = 9/2`,

`k = 3(9/2 - 6)(9/2 - 3)`
`k = 3(-3/2)(3/2)`
`k = -27/4`

Therefore the vertex is `(9/2, -27/4)`. Now plot the vertex and x-intercepts and join them with a free hand curve

Graph `y = -2(x - 4)(x - 5)`

Comparing the equation with intercept form `y = a(x - p)(x - q)`, p = 4 and q = 5. So the x-intercepts are `(4, 0)` and `(5, 0)`.
Graph `y = -2(x - 4)(x - 5)`

Let the vertex be (h, k), then `h = (p + q)/2 = (4 + 5)/2 = 9/2`,

`k = -2(9/2 - 4)(9/2 - 5)`
`k = -2(1/2)(-1/2)`
`k = 1/2`

Therefore the vertex is `(9/2, 1/2)`. Get some more points on the parabola by making table of x-y values

x
Y = -2(x - 4)(x - 5)
2
-12
3
-4
6
-4
7
-12

Now plot the vertex and x-intercepts and join them with a free hand curve

Graph `y = (x + 2)(x + 3)`

Comparing the equation with intercept form `y = a(x - p)(x - q)`, p = -2 and q = -3. So the x-intercepts are `(-2, 0)` and `(-3, 0)`.

Let the vertex be (h, k), then `h = (p + q)/2 = (-2 + -3)/2 = -5/2`,

`k = (-5/2 + 2)(-5/2 + 3)`
`k = (-1/2)(1/2)`
`k = -1/4`

Therefore the vertex is `(-5/2, -1/4)`. Now plot the vertex and x-intercepts and join them with a free hand curve.  Get some more points on the parabola by making table of x-y values

x
y = (x + 2)(x + 3)
-1
2
-4
2
0
6
-5
6
Now plot the vertex and x-intercepts and join them with a free hand curve
Graph of `y = (x + 2)(x + 3)`
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Graphing quadratics in Intercept Form

Instructions

The intercept form of a quadratic equation is y = a(x - p)(x - q), where 'p' and 'q' are the x-intercepts and 'a' is the value that determines the vertical stretch and whether the parabola is upside down.

Now since you already have the x-intercepts of the parabola, you already know two of its points. The other point which you must know is the vertex.

In order to get the the vertex, follow the method described below:

  • Let the vertex be (h, k)
  • h = (p + q)/2
  • k = f(h)
In the above method, 'h' is the x coordinate of the vertex and 'k' is the y coordinate of the vertex. In order to get the value of 'h', use the formula `h = (p + q)/2` and in order to get the value of 'k', substitute the value of h in place of x in the quadratic equation.

After getting the vertex, now you have three points on the parabola, its vertex and two x-intercepts. Plot these points on a coordinate plane and join them with the help of a free hand curve. In order to make the curve more accurate or larger, you can get more points on the parabola by direct substition method (that is making a T chart)

Example

Graph the parabola y = 4(x - 1)(x - 3)

On comparing with y = a(x - p)(x - q), 
  • a = 4
  • p = 1
  • q = 3
So the x-intercepts are (1, 0) and (3, 0).
Let the vertex be (h, k), then 
`h = (p + q)/2 = (1 + 3)/2 = 2`
`k = 4(h - 1)(h - 3) = 4(2 - 1)(2 - 3) = 4(1)(-1) = -4`
The vertex is (2, 4)

Now plot the two x-intercepts and vertex on a coordinate plane and join them with the help of a free hand curve.

Note that you can get more points by direct substitution method in order to make the curve more accurate. This can be done by taking x values and plugging them in the equation as follows:

`x`
`y = 4(x - 1)(x - 3)`
0
12
0.5
a
1.5
-3
2.5
-3
4
12
Note that you can take any value of x in the above table. These are only some example values.

Graph `y = 4(x - 1)(x - 3)`

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