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.

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.

**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**

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

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)**

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**

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