Chapter 3: Problem 1
Find the zeros of the following functions graphically: (a) \(f(x)=x^{2}-8 x+15\) (b) \(g(x)=2 x^{2}-4 x-16\)
Short Answer
Expert verified
The zeros of \( f(x) = x^2 - 8x + 15 \) are \( x = 3 \) and \( x = 5 \), and the zeros of \( g(x) = 2x^2 - 4x - 16 \) are \( x = -2 \) and \( x = 4 \).
Step by step solution
01
Understand the Problem
The problem requires finding the zeros of two quadratic functions graphically. Zeros of a function occur where the graph intersects the x-axis.
02
Graph the Function f(x)
Graph the function \( f(x) = x^2 - 8x + 15 \). This is a parabola opening upwards. Find the points where the parabola crosses the x-axis to determine the zeros.
03
Identify Zeros of f(x)
Upon graphing \( f(x) = x^2 - 8x + 15 \), the x-axis intersections occur at \( x = 3 \) and \( x = 5 \). Therefore, the zeros of \( f(x) \) are \( x = 3 \) and \( x = 5 \).
04
Graph the Function g(x)
Graph the function \( g(x) = 2x^2 - 4x - 16 \). This is also a parabola opening upwards. Determine where this parabola intersects the x-axis.
05
Identify Zeros of g(x)
Upon graphing \( g(x) = 2x^2 - 4x - 16 \), the parabola intersects the x-axis at \( x = -2 \) and \( x = 4 \). Thus, the zeros of \( g(x) \) are \( x = -2 \) and \( x = 4 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Zeros of a Function
When we talk about the zeros of a function, we refer to the values of the variable that make the function equal to zero. In simpler terms, it's where the graph of the function meets or touches the x-axis. For a quadratic function such as \( f(x) = ax^2 + bx + c \), the zeros can be found by determining the points where \( f(x) = 0 \).
- These points are crucial as they give insight into the roots or solutions of the quadratic equation.
- In exercises involving quadratic functions, finding the zeros graphically provides a visual representation of the concept.
Graphing Parabolas
Graphing a parabola involves plotting points that define the curving U-shape, characteristic of quadratic functions. The general form of a quadratic function is \( f(x) = ax^2 + bx + c \). Here's how to graph a parabola step-by-step:
- Determine the direction: If \( a > 0 \), the parabola opens upwards. If \( a < 0 \), it opens downwards.
- Find the vertex: The vertex is the highest or lowest point of the parabola, depending on its orientation. It can be calculated using the formula \((h, k)\), where \( h = -\frac{b}{2a} \) and \( k = f(h) \).
- Identify the symmetry: Parabolas are symmetrical about the vertical line known as the axis of symmetry, which passes through the vertex.
- Locate the zeros: Use the quadratic formula or factorization to find where the parabola intersects the x-axis.
X-Axis Intersections
Intersections with the x-axis are essential points on the graph of any function, representing where the function's output is zero. For quadratic functions, these intersections are synonymous with the zeros of the function.
- These intersections are vital because they represent the solutions to the equation \( ax^2 + bx + c = 0 \).
- If a quadratic function has two distinct real zeros, its graph will intersect the x-axis at two points. If there is one real zero, the graph just touches the x-axis, which is called a repeated or double root.
- If the quadratic has no real zeros, it means the parabola does not intersect the x-axis but rather lies entirely above or below it.
Solving Quadratic Equations
Solving quadratic equations means finding the values of \( x \) that satisfy the equation \( ax^2 + bx + c = 0 \). There are several methods to solve these equations, each with its strengths:
- Factoring: When possible, factoring the quadratic into simpler binomials makes solving straightforward as it leads directly to the zeros.
- Quadratic Formula: Used universally for all quadratics and given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), it always provides the solutions (real or complex).
- Completing the Square: This method involves reshaping the quadratic into a perfect square trinomial, then solving traditionally.