Question

Given that \(f(x)=5 x^{2}+x-7\) determine: (i) \(f(2) \div f(1)\) (ii) \(f(3+a)\) (iii) \(f(3+a)-f(3)\) (iv) \(\frac{f(3+a)-f(3)}{a}\)

Short answer

(i) -15, (ii) 5a^2 + 31a + 41, (iii) 5a^2 + 31a, (iv) 5a + 31

Step-by-step solution

Evaluate f(2)

Substitute \( x = 2 \) into the function \( f(x) = 5x^2 + x - 7 \). Calculate: \[f(2) = 5(2)^2 + (2) - 7 = 20 + 2 - 7 = 15\]

Evaluate f(1)

Substitute \( x = 1 \) into the function \( f(x) = 5x^2 + x - 7 \). Calculate: \[f(1) = 5(1)^2 + (1) - 7 = 5 + 1 - 7 = -1\]

Calculate f(2) ÷ f(1)

Use the values from Step 1 and Step 2. Calculate: \[f(2) \, \div \, f(1) = \frac{15}{-1} = -15\]

Substitute 3+a into the Function

Replace \( x \) in \( f(x) = 5x^2 + x - 7 \) with \( 3+a \) and simplify:\[f(3+a) = 5(3+a)^2 + (3+a) - 7\]Expand \( (3+a)^2 \):\[= 5(9 + 6a + a^2) + 3 + a - 7\]Distribute and simplify:\[= 45 + 30a + 5a^2 + 3 + a - 7 = 5a^2 + 31a + 41\]

Calculate f(3)

Substitute \( x = 3 \) into the function \( f(x) = 5x^2 + x - 7 \) and simplify:\[f(3) = 5(3)^2 + 3 - 7 = 45 + 3 - 7 = 41\]

Calculate f(3+a) - f(3)

Use the expressions from Step 4 and Step 5. Subtract the two:\[f(3+a) - f(3) = (5a^2 + 31a + 41) - 41 = 5a^2 + 31a\]

Simplify \\(\\frac{f(3+a) - f(3)}{a}\\)

Divide the expression obtained in Step 6 by \( a \):\[\frac{f(3+a) - f(3)}{a} = \frac{5a^2 + 31a}{a} = 5a + 31\] Simplify by canceling the \( a \) from numerator and denominator.

Key concepts

Polynomial Functions

Polynomial functions are a type of function that includes terms built from variables raised to various powers, added or subtracted together. They look like this:

• Each term has a coefficient and a variable raised to a non-negative integer power.
• The function in this exercise is a quadratic polynomial, meaning its highest power is a square, as in \(5x^2 + x - 7\).

A polynomial can have one, two, three, or more terms. Here, the function has three terms:

• The first term is \(5x^2\), which is called a quadratic term.
• The second term is \(x\), the linear term, and lastly,
• \(-7\), the constant term.

Each of these components gives us a full picture of the polynomial, defining its overall shape and behavior. When working with these functions, understanding the role of each term can be crucial in interpreting or manipulating the function in various contexts, such as evaluating values or performing transformations.

Substitution Method

The substitution method is a technique used to evaluate polynomial functions. Essentially, you plug a number into the function wherever you see the variable. Let's walk through the steps of this process in the given exercise:

• Step 1 was to find \(f(2)\): Substitute \(x = 2\) into the function. This substitution changes each instance of \(x\) in the equation to 2, leading to a straightforward calculation to evaluate the function value.
• Similarly, for \(f(1)\), we put \(x = 1\) into the function \(5x^2 + x - 7\).
• To find \(f(3+a)\), replace \(x\) with \(3+a\), which involves a bit more algebra as it requires expanding and simplifying.

This method is especially useful because it directly determines the output of the function for any given input, whether it is a constant or an algebraic expression, making it versatile for various types of problems.

Simplification

Simplification is the process of reducing a mathematical expression to its most concise form without changing its value. This operation requires understanding and applying algebraic rules and properties. Here’s how it was used in this exercise:

• For the expression \(f(3+a) = 5(3+a)^2 + (3+a) - 7\), expand the square and combine all like terms. This reduces the expression to a clearer form: \(5a^2 + 31a + 41\).
• We then used this to calculate \(f(3+a) - f(3)\), needing further simplification to \(5a^2 + 31a\).
• Finally, simplifying \(\frac{f(3+a) - f(3)}{a}\) involved canceling out the common factor of \(a\) in the numerator and denominator, leaving us with \(5a + 31\).

Simplification is about looking for these opportunities to make expressions smaller and clearer. It is an essential skill that not only makes calculations easier but also helps in accurately solving problems without unnecessary complexity.