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Consider the following quadratic function. Convert the function into standard form. f(x) = 4x^2 – 24x + 31

Consider the polynomial given below. Find the leading coefficient. P(x) = (x – 1)(x – 4)(x – 3)(x – ...

Consider the rational function f given below. Find the domain of f. (Enter your answer using interval notation.) f(x) = (x^3 + 27x^2 – 9)

Solve the inequality. Express your answer in interval form. (If there is no solution, enter NO SOLUTION.) 5x + 4 ≤ 3x^2

Solve the rational inequality. Express your answer using interval notation. (If there is no solution, enter NO SOLUTION.)

Paper For Above instruction

The assignment involves a series of mathematical problems focused on quadratic functions, polynomials, rational functions, and inequalities. These problems require transforming quadratic functions into standard form, identifying leading coefficients, determining the domain of rational functions, and solving both polynomial and rational inequalities. Mastery of algebraic manipulation, factoring, and interval notation is essential for successfully addressing these questions.

Introduction

Mathematics, particularly algebra, provides foundational tools for solving a variety of real-world problems through the study of functions and inequalities. Converting quadratic functions into standard form enables easier analysis of their graphs and properties. Recognizing key features like the vertex and intercepts becomes straightforward once the quadratic is in this form. Polynomial degree and coefficients, such as the leading coefficient, help determine end behavior and graph shape, which are essential for understanding polynomial functions. Rational functions, involving ratios of polynomials, require careful domain analysis because the denominator cannot be zero. Solving inequalities extends these concepts into the realm of solution sets, which are often expressed using interval notation for clarity and precision.

Converting a Quadratic Function to Standard Form

The given quadratic function is f(x) = 4x^2 – 24x + 31. To convert this into standard form, we complete the square. First, factor out the coefficient of x^2 from the quadratic terms:

f(x) = 4(x^2 – 6x) + 31

Next, complete the square inside the parentheses. Take half of the coefficient of x (which is -6), square it, and add inside the parentheses:

Half of -6 is -3, and (-3)^2 = 9

Add and subtract 9 inside the parentheses (note that adding 9 inside the parentheses is equivalent to adding 36 outside, because of the factor 4):

f(x) = 4(x^2 – 6x + 9 – 9) + 31 = 4((x – 3)^2 – 9) + 31

Distribute the 4:

f(x) = 4(x – 3)^2 – 36 + 31 = 4(x – 3)^2 – 5

Thus, the quadratic function in standard form is:

f(x) = 4(x – 3)^2 – 5

Identifying the Leading Coefficient of a Polynomial

The polynomial P(x) = (x – 1)(x – 4)(x – 3)(x – ...) appears to be a product of linear factors. The degree of the polynomial is 4, and the leading coefficient is the coefficient of the term with the highest degree, which is determined by multiplying the coefficients of the leading terms of each factor. Each linear factor (x – a) has a leading coefficient of 1; multiplying these together yields a leading coefficient of 1 for the polynomial. Therefore, the leading coefficient of P(x) is 1.

Finding the Domain of a Rational Function

Given the rational function f(x) = (x^3 + 27x^2 – 9), the domain is all real numbers except where the denominator equals zero. However, the provided function appears incomplete in the prompt; assuming it should be in the form f(x) = P(x)/Q(x), and based on context, we interpret the problem as finding the domain of a rational function where the denominator is a polynomial involving x. If the denominator is not explicitly provided, typically, the domain excludes points where the denominator is zero. Assuming the function is f(x) = (x^3 + 27x^2 – 9)/Q(x), the domain is all real numbers except solutions to Q(x) = 0. Without the denominator expression, we cannot explicitly solve, but the procedure involves setting the denominator equal to zero and solving for x to exclude those points.

Solving Polynomial Inequalities

The inequality 5x + 4 ≤ 3x^2 can be rewritten as 3x^2 – 5x – 4 ≥ 0 to standardize the form. Factoring or using the quadratic formula to find critical points:

Discriminant: Δ = (–5)^2 – 4 3 (–4) = 25 + 48 = 73

The roots are:

• x = [5 ± √73] / (2 * 3) = [5 ± √73] / 6

Since the quadratic coefficient is positive (3), the parabola opens upward, so the inequality 3x^2 – 5x – 4 ≥ 0 holds where x ≤ [5 – √73]/6 or x ≥ [5 + √73]/6. The solution set in interval notation is:

(–∞, [5 – √73]/6] ∪ [ [5 + √73]/6, ∞)

Solving Rational Inequalities

For the rational inequality 0 for all real x, the fraction 1 / (x^2 + 16) is always positive. Therefore, the inequality has no solution, resulting in NO SOLUTION.

Conclusion

This set of problems highlights essential algebraic techniques such as completing the square, analyzing polynomial coefficients, determining domain restrictions for rational functions, and solving inequalities. Mastery of these concepts is crucial for understanding the behavior of algebraic functions and their solution sets, which form the foundation for more advanced mathematics and real-world applications.

References

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