Question

Use the function f(x) to answer the questions:

f(x) = 2x2 − 3x − 5

Part A: What are the x-intercepts of the graph of f(x)? Show your work. (2 points)

Part B: Is the vertex of the graph of f(x) going to be a maximum or a minimum? What are the coordinates of the vertex? Justify your answers and show your work. (3 points)

Part C: What are the steps you would use to graph f(x)? Justify that you can use the answers obtained in Part A and Part B to draw the graph. (5 points)

Answer 1

To analyze the function \( f(x) = 2x^2 - 3x - 5 \), we will address each part of the question step by step.

### Part A: Finding the x-intercepts

To find the x-intercepts, we set \( f(x) = 0 \):

\[
2x^2 - 3x - 5 = 0
\]

We can solve this quadratic equation using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \( a = 2 \), \( b = -3 \), and \( c = -5 \).

First, we calculate the discriminant (\( b^2 - 4ac \)):

\[
b^2 - 4ac = (-3)^2 - 4(2)(-5) = 9 + 40 = 49
\]

Now we can substitute back into the quadratic formula:

\[
x = \frac{-(-3) \pm \sqrt{49}}{2 \cdot 2}
\]
\[
x = \frac{3 \pm 7}{4}
\]

This gives us two solutions:

1. \( x = \frac{3 + 7}{4} = \frac{10}{4} = \frac{5}{2} \)
2. \( x = \frac{3 - 7}{4} = \frac{-4}{4} = -1 \)

Thus, the x-intercepts of the graph of \( f(x) \) are:

\[
\left(-1, 0\right) \quad \text{and} \quad \left(\frac{5}{2}, 0\right)
\]

### Part B: Determining the vertex and its nature

The vertex of a parabola described by the function \( f(x) = ax^2 + bx + c \) occurs at the x-coordinate given by:

\[
x = -\frac{b}{2a}
\]

Substituting \( a = 2 \) and \( b = -3 \):

\[
x = -\frac{-3}{2 \cdot 2} = \frac{3}{4}
\]

Next, we find the corresponding y-coordinate by substituting \( x = \frac{3}{4} \) back into the function \( f(x) \):

\[
f\left(\frac{3}{4}\right) = 2\left(\frac{3}{4}\right)^2 - 3\left(\frac{3}{4}\right) - 5
\]
\[
= 2 \cdot \frac{9}{16} - \frac{9}{4} - 5
\]
\[
= \frac{18}{16} - \frac{36}{16} - \frac{80}{16}
\]
\[
= \frac{18 - 36 - 80}{16} = \frac{-98}{16} = -\frac{49}{8}
\]

Thus, the coordinates of the vertex are:

\[
\left(\frac{3}{4}, -\frac{49}{8}\right)
\]

Since the coefficient \( a = 2 \) is positive, the parabola opens upwards. Therefore, the vertex represents a **minimum** point.

### Part C: Steps to graph \( f(x) \)

To graph the function \( f(x) = 2x^2 - 3x - 5 \), follow these steps:

1. **Identify the x-intercepts**: From Part A, we have the x-intercepts at \( (-1, 0) \) and \( \left(\frac{5}{2}, 0\right) \).
2. **Identify the vertex**: From Part B, we found the vertex at \( \left(\frac{3}{4}, -\frac{49}{8}\right) \), which is a minimum point.
3. **Plot the vertex** and x-intercepts on the graph.
4. **Determine the axis of symmetry**: The graph is symmetric around the line \( x = \frac{3}{4} \).
5. **Select additional points** if necessary: Choose points on either side of the vertex and calculate their corresponding y-values to better understand the shape of the parabola.
6. **Draw the graph**: Sketch the parabola using the vertex, x-intercepts, and additional points, making sure it opens upwards.

We can use the x-intercepts and the vertex to properly establish the shape and direction of the graph. The x-intercepts give us the points where the graph crosses the x-axis, while the vertex determines the lowest point of the graph, confirming it opens upward based on our findings in Part B.

Therefore, we can accurately depict the graph of \( f(x) \) using this information.