Problem 19 - Algebra 2

Question

Find \(p(4)\) and \(p(-2)\) for each function. \(p(x)=x^{2}-3 x+8\)

Step-by-Step Solution

Answer

\(p(4) = 12\), \(p(-2) = 18\).

Step 1: Understand the Problem

We need to evaluate the function \(p(x) = x^2 - 3x + 8\) at two specific values of \(x\), namely \(x = 4\) and \(x = -2\). This involves substituting these values into the function to find \(p(4)\) and \(p(-2)\).

Step 2: Substitute x = 4 into the Function

Substitute 4 for \(x\) in \(p(x) = x^2 - 3x + 8\): \[ p(4) = (4)^2 - 3(4) + 8 \]Calculate: \[ p(4) = 16 - 12 + 8 = 12 \]

Step 3: Substitute x = -2 into the Function

Next, substitute -2 for \(x\) in \(p(x) = x^2 - 3x + 8\): \[ p(-2) = (-2)^2 - 3(-2) + 8 \]Calculate: \[ p(-2) = 4 + 6 + 8 = 18 \]

Step 4: Summarize the Results

After performing the calculations, we find that:- \(p(4) = 12\)- \(p(-2) = 18\).This concludes our evaluation for the given function at the specified points.

Key Concepts

Substitution Method

The substitution method is a fundamental algebraic technique used to solve equations or evaluate expressions by replacing variables with specific values. In this context, we use it to evaluate the polynomial function at given values of the variable. Let's break it down:

• Identify the function or expression where substitution is needed. Here, it's the polynomial function \( p(x) = x^2 - 3x + 8 \).


• Select the values to substitute into the function. In this exercise, we are given \( x = 4 \) and \( x = -2 \).


• Substitute each value into the expression, replacing \( x \) with the chosen number. Calculate to find the result.

By systematically substituting each specific value into our equation, we can determine what outputs the function yields. In this problem, substituting \( x = 4 \) leads to \( p(4) = 12 \), while substituting \( x = -2 \) results in \( p(-2) = 18 \). This method is crucial in understanding and analyzing functions, especially in checking how a function behaves at various points.

Quadratic Function

Quadratic functions are a type of polynomial function characterized by their highest degree being two. They generally appear in the form \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( a \) is non-zero. Let's see some important features of quadratic functions:

• They graph as a parabola, which can open upwards or downwards depending on the sign of \( a \).


• The vertex is the highest or lowest point of the parabola, determined using the formula \( (-\frac{b}{2a}, f(-\frac{b}{2a})) \).


• The axis of symmetry is the vertical line \( x = -\frac{b}{2a} \), dividing the parabola into two mirrored halves.

In the function \( p(x) = x^2 - 3x + 8 \) from the exercise, \( a = 1 \), \( b = -3 \), and \( c = 8 \). This means the parabola opens upward since \( a \) is positive. Understanding these features can help you visualize and analyze quadratic functions more effectively.

Algebra 2

Algebra 2 extends the foundations laid in Algebra 1 by exploring more complex mathematical concepts, including polynomial expressions, their functions, and operations. It builds upon basic algebraic skills to enable problem-solving in various mathematical scenarios. Key elements of Algebra 2 include:

• Understanding and working with different forms of polynomials.


• Solving quadratic equations using methods such as factoring, completing the square, and the quadratic formula.


• Manipulating expressions to simplify or rewrite them in an equivalent form.

In the provided exercise, students are practicing polynomial evaluation—a core concept in Algebra 2—using the substitution method. This step helps solidify understanding of evaluating expressions. By evaluating the quadratic polynomial at specific x-values, students become more comfortable working with these types of functions, preparing them for more advanced algebraic concepts and applications.