# Precalculus Assignment Help

homework assignment help is most useful online portal for students providing all type of Online Precalculus assignment help Services **Precalculus Assignment Help**, (or Algebra 3 in some areas) an advanced form of secondary school algebra, is a foundational mathematical discipline. It is also called Introduction to Analysis. In many schools, precalculus is actually two separate courses: Algebra and Trigonometry. Precalculus prepares students for calculus the same way as pre-algebra prepares students for Algebra I. While pre-algebra teaches students many different fundamental algebra topics, precalculus does not involve calculus, but explores topics that will be applied in calculus. Some precalculus courses might differ with others in terms of content. For example, an honors level course might spend more time on topics such as conic sections, vectors, and other topics needed for calculus. A lower level class might focus on topics used in a wider selection of higher mathematical areas, such as matrices which are used in business.

## Introduction to Pre calculus:

Pre calculus is superior form of secondary algebra, is a foundational mathematical discipline of American mathematics education, It is also called Introduction to Analysis. In many schools, pre calculus is actually contains two separate classes Algebra and Trigonometry. Pre calculus does not prepare students for calculus as pre-algebra is only for prepares students. Pre-algebra contains many different fundamental algebra topics; pre calculus does not involve calculus, but explores topics that will be applied in calculus. Some pre calculus courses are mostly differing from others in terms of content.

## Examples Problems on Pre Calculus:

**Example 1:**

Calculate h (4) and g(4) and h(4) / g(4) where functions g and h are defined by

h (x) = 3x - 8 and g (x) = x^{ 2} - 16

**Solution:**

Calculate h(4)

h(4) = 3(4) - 8 = 4

Calculate g (4)

g (4) = 4 ^{ 2} - 16

= 16 -16 = 0

By calculating h (4) / g(4) that g(4) in the denominator is equal to 0. The zero in the denominator defines undefined**h (4) / g (4) = undefined**

### Example 2:

Calculate the equation of quadratic function f whose graph has x intercepts (1, 0) and (-2, 0) and passes through the point (2,-8).

### Solution:

While the graph has x intercepts at the points (1, 0) and (-2, 0) and the equation may be written as follows.

f(x) = a(x - 1)(x + 2)

Function f passes through the point (2,-8) it follows that

f(2) = -8

Which leads to

-8 = a (2 - 1) (2 + 2)

Rewrite the right hand side if the equation and solve it.

-8 = 4a

Simplify the above equation for a to obtain

a = -2

The equation of function f is given by **f(x) = -2(x - 1) (x + 2)**

## Pre Calculus Online Practice Problem:

1) Simplify the polynomial equation. (x + 2)^{ 2} – 1

**Answer: (x + 3)(x + 1)**

2) Simplify the polynomial equation. x^{ 3} + x^{ 2} - 4x – 4

**Answer: (x + 1) (x + 2)(x - 2)**

3) Calculate h (4), g (4) and h (4) / g (4) where functions h and g are defined by

h (x) = 4x - 6 and g (x) = x^{ 2} - 25

**Answer: h (4) / g (4) = -10/9**

## Example Problemsfor Precalculus Practice

**Precalculus practice problem 1:**

Find the roots for the given quadratic equation 3x^{2} + 11x + 10.

**Solution:**

Given 3x^{2 }+ 11x + 10

Factorize the given equation, we get

3x^{2 }+ 11x + 10 = 3x^{2} + 6x + 5x + 10

= 3x (x + 2) + 5 (x + 2)

= (3x + 5) (x + 2)

Equating the equation to zero, we get

(3x + 5) (x + 2) = 0.

3x + 5 = 0, x + 2 = 0.

x = (- 5 / 3), x = - 2.

After solving, we get

The roots are (- 5 / 3), - 2.

**Answer:**

The roots are x = (- 5 / 3) and x = - 2.

**Precalculus **** practice problem ****2:**

Solve the given factorial (`(6!) / (4!)`).

**Solution:**

Given (`(6!) / (4!)`)

**Formula: **

n! = n (n - 1) (n - 2).... .1.

(`(6!) / (4!)`) = `(6!) / (4!)`

6! = 6 * 5 * 4 * 3* 2 *1

= 720.

4! = 4 * 3 * 2 * 1

= 24

Therefore, we get

`(6!) / (4!)` = `(720) / (24)`

After solving, we get

** **= 30

**Answer:**

The final answer `(6!) / (4!)` = 30

**Precalculus ****practice problem**** 3:**

Find the slope and equation of the straight line, it passes through the points (0,4) and (4, 6).

**Solution:**

Given points are (0,6) and (4, 10)

Here, x_{1} = 0, x_{2} = 4, y_{1} = 6, and y_{2} = 10

**Formula for finding slope:**

Slope (m) = `(y_2 - y_1) / (x_2 - x_1)`

Substitute the given values in the above formula, we get

= `(10 - 6) / (4 - 0)`

m = 1

**Formula for line equation:**

(y - y_{1}) = m (x - x_{1})

Substitute the given values, we get

y - 6 = 1 ( x - 4)

y - 6 = x - 4

Add 6 on both the sides, we get

y = x + 2

**Answer:**

Slope of the equation is (m) = 1

Line equation is y = x + 2

## Practice Problems for Precalculus

**Precalculus practice problem 1:**

Solving the given factorial and find its value (`(12!) / (10!)`)

**Answer:**

The final answer is 132

**Precalculus practice problem 2:**

Find the slope of the given straight line passes through the points (7, 8) and (9, 16)

**Answer:**

The final answer is Slope m = 4

**Precalculus practice problem 3:**

Multiply the given expressions (2x + 17) (x^{2} - 15)

**Answer:**

The final answer is 2x^{3} - 17x^{2} - 30x - 255

** precalculus preparation:**

Precalculus is an advanced form of secondary school algebra, is a foundational mathematical discipline. It is also called Introduction to Analysis. In many schools, precalculus is the actually two separate courses: Algebra and Trigonometry. Precalculus preparation does not prepare students for the calculus as pre-algebra prepares students for Algebra. While pre-algebra teaches student many fundamental algebra topics, precalculus does not involve calculus, but explores topics that will be applied in calculus.

## Preparation for Precalculus :

**Preparation for Complex Numbers**

**Example 1:**

Multiply (5+ 2i) and (2 + 5i)

**Solution :**

Given problem is in the form of (a+bi) × (a+bi)

(5 + 2i)(2 + 5i) = (5× 2) + (5 × (5i)) + ((2i) × 2) + ((2i) × (5i))

= 10+ 25i + 4i + 10i² Where i^{2}=-1

= 10 + 29i -10 = 10(-1) = -10

= 0 + 29i

=29i

Therefore, **(2 + 2i)(5+ 5i) = 29i**

**Example 2:**

Find the complex conjugate of 2 + i 7

**Solution :**

By definition, the complex conjugate is obtained by reversing the sign of the imaginary part of the complex number.Hence the required conjugates are 2 − i 7,

**preparation for Trigonometric Functions**

In a triangle ABC, we have that sinA = 7/9, what are the values of cosA and tanA?

**Solution:**

We have the basic trigonometric identity:

sin^{2 }A + cos^{2} A = 1

In our case, we know that sin A = 7/9, so we replace in the previous equation to get:

cos^{2} A = 1-(7/9)^{2}

We want to solve for cosA so we get

cos^{2} A = 1 −(49/81)=32/81

so we apply square root to get

cos A =sqrt(32/81)=4√2/9.

To find tan A we recall that the tan A = sin A / cos A, so using the information we already have we find that

Tan A =(7/9)/(4√2/9) = 7/ 4√2= 7√2/8

Tan A=7√2/8

## Practice Problem Preparation for Precalculus:

If A, B, C, D are angles of a cyclic quadrilateral prove that cosA + cosB + cos C + cos D = 0.

Find the real and imaginary parts of the complex number z =3i

^{20}− i^{19}/ 2i − 1

** Answer:*** Re*(*z*) = −1/5 and *Im*(*z*) =− 7 / 5

** Solving precalculus:**

## Solving Precalculus Problem:

**Solving Inverse function:**

**Example:**

Show that the function y = x^{2} is not one-to-one and inverse contitions

**Solution:**

For the different values of x (say 1, − 1) we have the same value of y. i.e. different elements in the domain have the same element in the co-domain. By definition of one-to-one, it is not one-to-one (OR)

y = f(x) = x^{2}

f(1) = 1^{2} = 1

f(− 1) = (− 1)^{2} = 1

⇒ f(1) = f(− 1)

But 1 ≠ − 1. Thus different objects in the domain have the same image.

∴ the function is not one-to-one.

** Solving Rational function:**

**Example:**

** **Find the domain of the rational function f(x) =x^{2} + x + 2 / x^{2} − x

**Solution:**

The domain S is obtained by removing all the points from R for which g(x)

= 0 ⇒ x^{2 }− x = 0 ⇒ x(x − 1) = 0 ⇒ x = 0, 1

∴ S = R − {0, 1}

Thus this rational function is defined for all real numbers except 0 and 1.

**Solving Composition of functions:**

**Example:**

The two function f : R → R, g : R → R are defined by f(x) = x^{2} + 1, g(x) = x − 1. Find fog and gof and show that fog ≠ gof.

**Solution:**

(fog) (x) = f(g(x)) = f(x − 1) = (x − 1)^{2} + 1 = x^{2} − 2x + 2

(gof) (x) = g(f(x)) = g(x^{2} + 1) = (x^{2} + 1) − 1 = x^{2}

Thus (fog) (x) = x^{2 }− 2x + 2

(gof) (x) = x^{2}

⇒ fog ≠ gof

## Practice Problem for Solving Pre Calculus:

1. Let f : R → R be a function defined by f(x) = 3x + 1. Find f ^{−1}

**Answer: **x - 1/3

2.Let A = {1, 2}, B = {3, 4} and C = {5, 6} and f : A → B andg : B → C such that f(1) = 3, f(2) = 4, g(3) = 5, g(4) = 6. Find gof.

**Answer:** gof = {(1, 5), (2, 6)}

** PRECALCULUS HELP :**

Precalculus help is a relationship between Algebra and Calculus. Precalculus help consists of deepness review of topics from Algebra; it presents a sensibly complete look at trigonometry, progressing from right triangle based definitions to functions of real numbers and their graphs. Precalculus help also deals with Trigonometric ratios, Hyperbolas, circles, limit of a function, etc.

## Trigonometric Identities on Precalculus Help

**Pythagorean Identities**

- sin
^{2}x + cos^{2}x = 1 - tan
^{2}x + 1 = sec^{2}x - 1 + cot
^{2}x = csc^{2}x

These are the precalculus help Identities we use mainly in solving problems.

## Examples of Precalculus

**Question 1: If a triangle ABC, we have that sin A = 7/9, what are the values of cos A and tan A?**

**Solution:**

We have the basic trigonometric identity:

sin^{2 }A + cos^{2} A = 1

In our case, we know sin A = 7/9, so we replace in the previous equation to get:

cos^{2} A = 1-`(7/9)` ^{2}

We want to solve for cos A so we get

cos^{2} A = 1 −`(49/81)` =`32/81`

So we apply square root to get

** Cos A =√` (32/81)` =4√`2/9.` **

To find tan A we recall that tan A = `sin A / cos A` , so using the information we already have we find that

Tan A = `(7/9)` /(4√`2/9` ) = `7/ 4` √2= 7√`2/8`

** Tan A=7√`2/8` **

**Therefore ** **Cos A =4√`2/9` and Tan A=7√`2/8` **

**Example 2:** Prove the identity cos x * tan x = sin x

Solution

- We start with the left side and change it into sin x.
- To Use tan x = sin x /cos x identity in the left side.

cos x * tan x = cos x *` (sin x / cos x) ` = sin x

**Example 3:** Prove the identity cot a * sec a * sin a = 1

**Solution:**

- Use the identities cot a = `cos a / sin a ` and sec a = `1/ cos a` in the left side.

cot a * sec a * sin a = `(cos a / sin a)` * `(1/ cos a)` * sin a - Simplify to obtain.

`(cos a / sin a)` * `(1/ cos a)` * sin a = 1

**Question 5: sin x (cot x + tan x) = sec x**

**LHS:**

sin x (`cos x / sin x` + `sin x / cos x` )

cos x + `(sin^2x) / cos x` = (distribute and simplify)

(`cos^2 x` / cos x + `sin^2 x`` / cos x` ) = (multiply by `cos / cos` )

`(cos^2 x + sin^2 x) / cos x ` = (combine into single fraction)

`1 / cos x` = (since `cos^2 + sin^2` = 1)

** sec x = sec x RHS**

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