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Chapter 8: Problem 20

For each quadratic function, (a) find the vertex and the axis of symmetry and (b) graph the function. $$f(x)=x^{2}+2 x-5$$

Short Answer

Expert verified
Vertex: \((-1, -6)\), Axis of Symmetry: \(x = -1\).

Step by step solution

01

Identify the coefficient values

In the quadratic function of the form \(f(x) = ax^2 + bx + c\), identify the values of \(a\), \(b\), and \(c\) from the given equation \(f(x) = x^2 + 2x - 5\). Here, \(a = 1\), \(b = 2\), and \(c = -5\).
02

Calculate the axis of symmetry

The axis of symmetry for a quadratic function is given by the formula \(x = -\frac{b}{2a}\). Substitute the values of \(b\) and \(a\):\(x = -\frac{2}{2(1)} = -1\). Thus, the axis of symmetry is \(x = -1\).
03

Find the vertex

The vertex lies on the axis of symmetry. To find the y-coordinate of the vertex, substitute \(x = -1\) into the original function \(f(x) = x^2 + 2x - 5\): \(f(-1) = (-1)^2 + 2(-1) - 5 = 1 - 2 - 5 = -6\). Thus, the vertex is \((-1, -6)\).
04

Graph the function

To graph the quadratic function \(f(x) = x^2 + 2x - 5\): 1. Plot the vertex \((-1, -6)\). 2. The parabola opens upwards (since \(a > 0\)). 3. Plot additional points by choosing x-values around the vertex, for example: - For \(x = -2\), \(f(-2) = (-2)^2 + 2(-2) - 5 = 4 - 4 - 5 = -5\) - For \(x = 0\), \(f(0) = 0^2 + 2(0) - 5 = -5\). 4. Draw a smooth curve through the vertex and these points.

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Key Concepts

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

vertex of a parabola
The vertex of a parabola is the point where the graph changes direction and reaches its maximum or minimum value. It is a crucial feature in graphing quadratic functions. For a quadratic function in standard form, \(f(x) = ax^2 + bx + c\), the vertex can be found using the formula: \(x = -\frac{b}{2a}\). Substitute this x-value back into the original equation to find the y-coordinate. This gives the vertex coordinates \((x, y)\). In our example, with a quadratic function \(f(x) = x^2 + 2x - 5\), we calculated the vertex as follows: \(-1, -6\). The x-coordinate is \(x = -1\) and substituting it back into the function gives \(f(-1) = -6\). So, the vertex is at \((-1, -6)\).
axis of symmetry
The axis of symmetry is a vertical line that divides the parabola into two mirror-image halves. It always passes through the vertex. For a quadratic function in the form \(f(x) = ax^2 + bx + c\), the axis of symmetry can be found using the formula: \(x = -\frac{b}{2a}\). For our function, \(f(x) = x^2 + 2x - 5\), the axis of symmetry is: \(x = -\frac{2}{2(1)} = -1\). This means the parabola will be symmetrical along the line \(x = -1\). This line helps in plotting the graph as it shows that points on either side of it will mirror each other.
graphing quadratic functions
Graphing quadratic functions involves a few key steps to ensure accuracy. Here鈥檚 a quick guide:
  • Find the vertex and axis of symmetry.
  • Determine whether the parabola opens upwards (if \(a > 0\)) or downwards (if \(a < 0\)).
  • Plot the vertex on the graph.
  • Choose additional x-values around the vertex to get more points. Substitute these x-values into the function to get corresponding y-values.
  • Plot these points and draw a smooth curve through them, ensuring it is symmetric about the axis of symmetry.
In our example, after finding the vertex \((-1, -6)\) and axis of symmetry \(x = -1\), we plotted additional points like \((-2, -5)\) and \((0, -5)\). The curve bows upward forming U-shaped parabola.
standard form of quadratic equation
The standard form of a quadratic equation is written as: \(ax^2 + bx + c = 0\). Here, \(a\), \(b\), and \(c\) are constants, and \(x\) represents the variable. This is the most familiar way to express a quadratic equation and is crucial for identifying and calculating properties like the vertex, axis of symmetry, and direction of the parabola. For example, \(f(x) = x^2 + 2x - 5\) is in standard form with \(a = 1\), \(b = 2\), and \(c = -5\). The coefficients \(a\) determines the opening direction of the parabola. If \(a > 0\), the parabola opens upwards. If \(a < 0\), it opens downwards. Here, since \(a = 1 > 0\), the parabola opens upwards.

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