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Chapter 8: Problem 24

Find the vertex of the graph of each quadratic function. Determine whether the graph opens upward or downward, find any intercepts, and sketch the graph. See Examples 1 through 4 . $$ f(x)=x^{2}+2 x-3 $$

Short Answer

Expert verified
Vertex: (-1, -4), Graph opens upward, X-intercepts: (-3, 0) and (1, 0), Y-intercept: (0, -3).

Step by step solution

01

Identify the Quadratic Function Components

The given quadratic function is \( f(x) = x^2 + 2x - 3 \). This is in the form \( ax^2 + bx + c \) where \( a = 1 \), \( b = 2 \), and \( c = -3 \).
02

Determine the Direction of the Graph Opening

Since \( a = 1 \), which is positive, the parabola opens upwards.
03

Calculate the Vertex

Use the vertex formula \( x = -\frac{b}{2a} \) to find the x-coordinate of the vertex. Substituting \( a = 1 \) and \( b = 2 \), we get \( x = -\frac{2}{2 \times 1} = -1 \). To find the y-coordinate, substitute \( x = -1 \) back into the function: \( f(-1) = (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4 \). Thus, the vertex is \((-1, -4)\).
04

Find the X-Intercepts

Set \( f(x) = 0 \) and solve for \( x \): \( x^2 + 2x - 3 = 0 \). Factor the quadratic as \( (x + 3)(x - 1) = 0 \). Setting each factor to zero gives the x-intercepts: \( x = -3 \) and \( x = 1 \).
05

Find the Y-Intercept

The y-intercept occurs where \( x = 0 \). Substitute \( x = 0 \) into the function: \( f(0) = 0^2 + 2 \times 0 - 3 = -3 \). So, the y-intercept is \( (0, -3) \).
06

Sketch the Graph

Plot the vertex \((-1, -4)\), the x-intercepts \((-3, 0)\) and \((1, 0)\), and the y-intercept \((0, -3)\). Connect these points with a smooth curve opening upward to sketch the parabola.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parabola
In mathematics, a parabola is a U-shaped curve that represents the graph of a quadratic function. Quadratic functions come in the form of \( f(x) = ax^2 + bx + c \). The constant \( a \) determines the direction in which the parabola opens.
If \( a > 0 \) (like our function where \( a = 1 \)), the parabola opens upwards. If \( a < 0 \), it opens downwards. Parabolas have a symmetrical shape and are defined by several key features including the vertex, axis of symmetry, x-intercepts, and y-intercept. Understanding these components helps in sketching the graph effectively.
Vertex
The vertex of a parabola is the point where the curve changes direction. For an upwards opening parabola, it's the lowest point, while for a downwards opening parabola, it's the highest. The vertex provides a point of symmetry for the parabola. To find the vertex of the quadratic function, you can use the formula \( x = -\frac{b}{2a} \). This calculates the x-coordinate of the vertex. In our example, substituting \( a = 1 \) and \( b = 2 \) gives us \( x = -1 \). To find the y-coordinate, substitute this x value back into the function:
  • \( f(-1) = (-1)^2 + 2(-1) - 3 \)
  • \( f(-1) = 1 - 2 - 3 \)
  • \( f(-1) = -4 \)
Thus, the vertex is \((-1, -4)\).
X-Intercepts
The x-intercepts, or roots of the quadratic function, are the points where the graph crosses the x-axis. At these points, the value of the function \( f(x) \) is zero. To find them, set the quadratic equation to zero: \( x^2 + 2x - 3 = 0 \). Factoring this gives us \((x + 3)(x - 1) = 0\). Solving these equations gives:
  • \( x + 3 = 0 \) leads to \( x = -3 \)
  • \( x - 1 = 0 \) gives \( x = 1 \)
So, the x-intercepts are \( (-3, 0) \) and \( (1, 0) \). These points are crucial for the accurate sketch of the parabola.
Y-Intercept
The y-intercept is where the parabola crosses the y-axis. For a quadratic function, this is simply the constant term \( c \) in the expression \( ax^2 + bx + c \). However, it can also be found by setting \( x = 0 \) and solving for \( f(x) \). For our function, substituting \( x = 0 \) gives:
  • \( f(0) = 0^2 + 2 \times 0 - 3 \)
  • \( f(0) = -3 \)
So the y-intercept is \( (0, -3) \). This point helps in plotting the curve correctly.
Graph Sketching
Graph sketching is an essential skill in visualizing quadratic functions. Start by plotting the key points: the vertex, x-intercepts, and y-intercept. In our case, plot the vertex \((-1, -4)\), the x-intercepts \((-3, 0)\) and \((1, 0)\), and the y-intercept \((0, -3)\). Once plotted, connect these points with a smooth curve ensuring the parabola opens upwards as determined by the positive \( a \). Remember:
  • The vertex is the turning point of the parabola.
  • X-intercepts are where the curve crosses the x-axis.
  • The y-intercept is where the curve crosses the y-axis.
Using symmetry can help ensure accuracy in your sketch, giving a complete visual representation of the quadratic function.

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Most popular questions from this chapter

Simplify each expression. See Section 7.5. $$ \frac{10-20 \sqrt{3}}{2} $$

Solve. See the Concept Check in this section. Which description of \(f(x)=5\left(x+\frac{1}{2}\right)^{2}+\frac{1}{2}\) is correct? $$ \begin{array}{|l|l|} \hline \text {Graph Opens} & {\text {Vertex}} \\ \hline \text { a. upward } & {\left(\frac{1}{2}, \frac{1}{2}\right)} \\ \hline \text { b. upward } & {\left(-\frac{1}{2}, \frac{1}{2}\right)} \\ \hline \text { c. downward } & {\left(\frac{1}{2},-\frac{1}{2}\right)} \\ \hline \text { d. downward } & {\left(-\frac{1}{2},-\frac{1}{2}\right)} \\ \hline \end{array} $$

If you are borrowing money at a rate of \(10 \%,\) would you prefer the interest to be simple or compound? Explain your answer.

Solve. See Example 5 Find two numbers whose sum is 60 and whose product is as large as possible. [Hint: Let \(x\) and \(60-x\) be the two positive numbers. Their product can be described by the function \(f(x)=x(60-x) \cdot]\)

Sketch the graph of each function. See Section 8.5. $$ f(x)=(x-3)^{2} $$
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