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Chapter 5: Problem 40

Complete parts a-c for each quadratic function. a. Find the \(y\) -intercept, the equation of the axis of symmetry, and the \(x\) -coordinate of the vertex. b. Make a table of values that includes the vertex. c. Use this information to graph the function. $$ f(x)=2 x^{2}+5 x $$

Short Answer

Expert verified
The y-intercept is 0. The axis of symmetry is \( x = -\frac{5}{4} \).

Step by step solution

01

Identify the Quadratic Function

The quadratic function we are given is \( f(x) = 2x^2 + 5x \). In standard form, a quadratic function is \( f(x) = ax^2 + bx + c \). Here, \( a = 2 \), \( b = 5 \), and \( c = 0 \).
02

Find the y-intercept

The \( y \)-intercept is the value of \( f(x) \) when \( x = 0 \). Substitute \( x = 0 \) into the equation: \[ f(0) = 2(0)^2 + 5(0) = 0. \] So, the \( y \)-intercept is \( 0 \).
03

Determine the Axis of Symmetry

The axis of symmetry for a quadratic function in standard form is given by the formula \( x = -\frac{b}{2a} \). Plug in the values for \( b \) and \( a \):\[ x = -\frac{5}{2(2)} = -\frac{5}{4}. \] Thus, the axis of symmetry is \( x = -\frac{5}{4} \).
04

Find the x-coordinate of the Vertex

The \( x \)-coordinate of the vertex is the same as the axis of symmetry, which is \( x = -\frac{5}{4} \).
05

Make a Table of Values

Choose values around the vertex \( x = -\frac{5}{4} \): Let's choose \( x \) values such as \(-2, -1, 0\), and identify \( f(x) \) for these. Calculate:- \( f(-2) = 2(-2)^2 + 5(-2) = 8 - 10 = -2. \)- \( f(-1) = 2(-1)^2 + 5(-1) = 2 - 5 = -3. \)- \( f(0) = 0. \)Include the vertex calculated at \( -\frac{5}{4} \), - Use \( f\left(-\frac{5}{4}\right) = 2\left(-\frac{5}{4}\right)^2 + 5\left(-\frac{5}{4}\right) = 2\left(\frac{25}{16}\right) - \frac{25}{4} = \frac{50}{16} - \frac{100}{16} = -\frac{50}{16} = -\frac{25}{8}. \)So, points include: (-2, -2), (-1, -3), \(- \frac{5}{4}, -\frac{25}{8}\), (0, 0)
06

Graph the Function

On a graph, plot the points calculated: (-2, -2), (-1, -3), \(-\frac{5}{4}, -\frac{25}{8} \), and (0, 0). Draw the parabola that passes through these points, ensuring it is symmetric about the line \( x = -\frac{5}{4} \).

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Key Concepts

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

Axis of Symmetry
In every quadratic function of the form \( ax^2 + bx + c \), the axis of symmetry is a vertical line that can be calculated using the formula \( x = -\frac{b}{2a} \). This line is crucial because it divides the parabola into two mirror-image halves. For the quadratic function \( f(x) = 2x^2 + 5x \), you can find the axis of symmetry by substituting \( a = 2 \) and \( b = 5 \) into the formula. This gives us \( x = -\frac{5}{4} \). The axis of symmetry helps in determining the vertex and guiding the overall shape of the parabola during graphing.
Vertex of a Parabola
The vertex of a parabola is the point where the curve is at its maximum or minimum. For the function \( f(x) = 2x^2 + 5x \), the vertex lies on the axis of symmetry. Once the \( x \) coordinate of the vertex is calculated, plugging it back into the quadratic function will give you the \( y \) coordinate. Here, the \( x \)-coordinate is \(-\frac{5}{4}\). Calculating the function at this point, \( f(-\frac{5}{4}) \), results in a \( y \)-coordinate of \(-\frac{25}{8}\). Therefore, the vertex of the parabola is \( \left(-\frac{5}{4}, -\frac{25}{8}\right) \). This particular vertex finds itself below the x-axis, indicating a downward direction of the curve.
Y-Intercept
The \( y \)-intercept of a quadratic is simply the point where the parabola crosses the \( y \)-axis. It is determined by evaluating the function at \( x = 0 \). For \( f(x) = 2x^2 + 5x \), substituting \( x = 0 \) results in a \( y \)-intercept at 0. This means that the point \( (0,0) \) is where the parabola touches the \( y \)-axis. Understanding the \( y \)-intercept is useful as it offers one fixed point on the graph of the function.
Table of Values
Creating a table of values helps in understanding the behavior of the quadratic function over a range of \( x \) values. It is particularly handy for plotting the function on a graph. For \( f(x) = 2x^2 + 5x \), select values of \( x \) around the vertex \(-\frac{5}{4}\). Points like \(-2, -1, 0\) give a broader picture:
  • \( f(-2) = -2 \)
  • \( f(-1) = -3 \)
  • \( f(0) = 0 \)
Include also the vertex \(-\frac{5}{4}, -\frac{25}{8}\). These values enable a visual representation of how the parabola curves and where it changes direction.
Graphing Quadratic Equations
Graphing the quadratic function is the process of visually representing it on a coordinate plane. Start by plotting the points obtained from your table of values. For the function \( f(x) = 2x^2 + 5x \), you will plot:
  • Point \((-2, -2)\)
  • Point \((-1, -3)\)
  • Vertex \(\left(-\frac{5}{4}, -\frac{25}{8}\right)\)
  • Point \((0, 0)\)
Connect these points with a smooth curve forming a parabola, ensuring that the curve is symmetric about the axis \( x = -\frac{5}{4} \). Observing the plotted points makes it easier to see the shape that the quadratic function takes and predict its behavior over further values.

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