Standard Form Practice worksheet with three quadratic function problems involving graph analysis and graphing.

Problem Analysis and Solution

The worksheet involves analyzing quadratic functions in standard form and graphing them. Let's solve each part step by step.

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#### Problem 1: Analyze the function \( f(x) = x^2 + 4x + 7 \)

##### Given Graph Characteristics:
- The graph is a parabola opening upwards.
- The vertex is the lowest point on the graph.
- The y-intercept is where the graph crosses the y-axis.
- The zeros are where the graph crosses the x-axis.

##### Step-by-Step Solution:

1. Domain:
- The domain of a quadratic function is all real numbers because there are no restrictions on the input \( x \).
- Answer: \( (-\infty, \infty) \)

2. Range:
- The parabola opens upwards, so the range starts at the y-coordinate of the vertex and extends to infinity.
- From the graph, the vertex appears to be at \( (x, y) = (-2, 3) \). Thus, the minimum value of \( f(x) \) is 3.
- Answer: \( [3, \infty) \)

3. Axis of Symmetry:
- The axis of symmetry for a quadratic function \( ax^2 + bx + c \) is given by \( x = -\frac{b}{2a} \).
- Here, \( a = 1 \) and \( b = 4 \):
\[
x = -\frac{4}{2(1)} = -2
\]
- Answer: \( x = -2 \)

4. Vertex:
- The vertex is the point where the parabola reaches its minimum value. From the graph, it appears to be at \( (-2, 3) \).
- Answer: \( (-2, 3) \)

5. Y-intercept:
- The y-intercept occurs when \( x = 0 \). Substitute \( x = 0 \) into the function:
\[
f(0) = 0^2 + 4(0) + 7 = 7
\]
- Answer: \( (0, 7) \)

6. Zero(s):
- The zeros are the x-values where the graph crosses the x-axis. From the graph, there are no x-intercepts (the parabola does not touch the x-axis).
- Answer: None

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#### Problem 2: Analyze the function \( y = -x^2 - 5x + 1 \)

##### Given Graph Characteristics:
- The graph is a parabola opening downwards.
- The vertex is the highest point on the graph.
- The y-intercept is where the graph crosses the y-axis.
- The zeros are where the graph crosses the x-axis.

##### Step-by-Step Solution:

1. Domain:
- The domain of a quadratic function is all real numbers.
- Answer: \( (-\infty, \infty) \)

2. Range:
- The parabola opens downwards, so the range ends at the y-coordinate of the vertex and extends to negative infinity.
- From the graph, the vertex appears to be at \( (x, y) = \left(-\frac{5}{2}, \frac{29}{4}\right) \). Thus, the maximum value of \( y \) is \( \frac{29}{4} \).
- Answer: \( \left(-\infty, \frac{29}{4}\right] \)

3. Axis of Symmetry:
- The axis of symmetry for a quadratic function \( ax^2 + bx + c \) is given by \( x = -\frac{b}{2a} \).
- Here, \( a = -1 \) and \( b = -5 \):
\[
x = -\frac{-5}{2(-1)} = -\frac{5}{2}
\]
- Answer: \( x = -\frac{5}{2} \)

4. Vertex:
- The vertex is the point where the parabola reaches its maximum value. From the graph, it appears to be at \( \left(-\frac{5}{2}, \frac{29}{4}\right) \).
- Answer: \( \left(-\frac{5}{2}, \frac{29}{4}\right) \)

5. Y-intercept:
- The y-intercept occurs when \( x = 0 \). Substitute \( x = 0 \) into the function:
\[
y = -(0)^2 - 5(0) + 1 = 1
\]
- Answer: \( (0, 1) \)

6. Zero(s):
- The zeros are the x-values where the graph crosses the x-axis. From the graph, the zeros appear to be approximately \( x \approx -5.2 \) and \( x \approx 0.2 \).
- Answer: \( x \approx -5.2 \) and \( x \approx 0.2 \)

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#### Problem 3: Graph the quadratic function \( f(x) = 2x^2 + x - 6 \)

##### Step-by-Step Solution:

1. Find the Vertex:
- The axis of symmetry is given by \( x = -\frac{b}{2a} \).
- Here, \( a = 2 \) and \( b = 1 \):
\[
x = -\frac{1}{2(2)} = -\frac{1}{4}
\]
- Substitute \( x = -\frac{1}{4} \) into the function to find the y-coordinate of the vertex:
\[
f\left(-\frac{1}{4}\right) = 2\left(-\frac{1}{4}\right)^2 + \left(-\frac{1}{4}\right) - 6
\]
\[
= 2\left(\frac{1}{16}\right) - \frac{1}{4} - 6
\]
\[
= \frac{2}{16} - \frac{4}{16} - \frac{96}{16}
\]
\[
= \frac{2 - 4 - 96}{16} = \frac{-98}{16} = -\frac{49}{8}
\]
- The vertex is \( \left(-\frac{1}{4}, -\frac{49}{8}\right) \).

2. Find the Y-intercept:
- The y-intercept occurs when \( x = 0 \):
\[
f(0) = 2(0)^2 + (0) - 6 = -6
\]
- The y-intercept is \( (0, -6) \).

3. Find the Zeros:
- Solve \( 2x^2 + x - 6 = 0 \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
- Here, \( a = 2 \), \( b = 1 \), and \( c = -6 \):
\[
x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-6)}}{2(2)}
\]
\[
= \frac{-1 \pm \sqrt{1 + 48}}{4}
\]
\[
= \frac{-1 \pm \sqrt{49}}{4}
\]
\[
= \frac{-1 \pm 7}{4}
\]
\[
x = \frac{-1 + 7}{4} = \frac{6}{4} = \frac{3}{2} \quad \text{and} \quad x = \frac{-1 - 7}{4} = \frac{-8}{4} = -2
\]
- The zeros are \( x = \frac{3}{2} \) and \( x = -2 \).

4. Plot Additional Points:
- Choose a few more x-values to plot points and sketch the parabola.
- For example:
\[
f(-1) = 2(-1)^2 + (-1) - 6 = 2 - 1 - 6 = -5
\]
\[
f(1) = 2(1)^2 + (1) - 6 = 2 + 1 - 6 = -3
\]

5. Sketch the Graph:
- Plot the vertex \( \left(-\frac{1}{4}, -\frac{49}{8}\right) \), the y-intercept \( (0, -6) \), and the zeros \( \left(\frac{3}{2}, 0\right) \) and \( (-2, 0) \).
- Use additional points to sketch the parabola.

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Final Answers:

1. For \( f(x) = x^2 + 4x + 7 \):
- Domain: \( (-\infty, \infty) \)
- Range: \( [3, \infty) \)
- Axis of Symmetry: \( x = -2 \)
- Vertex: \( (-2, 3) \)
- Y-intercept: \( (0, 7) \)
- Zero(s): None

2. For \( y = -x^2 - 5x + 1 \):
- Domain: \( (-\infty, \infty) \)
- Range: \( \left(-\infty, \frac{29}{4}\right] \)
- Axis of Symmetry: \( x = -\frac{5}{2} \)
- Vertex: \( \left(-\frac{5}{2}, \frac{29}{4}\right) \)
- Y-intercept: \( (0, 1) \)
- Zero(s): \( x \approx -5.2 \) and \( x \approx 0.2 \)

3. Graph \( f(x) = 2x^2 + x - 6 \):
- Vertex: \( \left(-\frac{1}{4}, -\frac{49}{8}\right) \)
- Y-intercept: \( (0, -6) \)
- Zeros: \( x = \frac{3}{2} \) and \( x = -2 \)
- Sketch the parabola using these points.

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Final Answer Box:
\[
\boxed{
\begin{array}{l}
\text{1. Domain: } (-\infty, \infty), \text{ Range: } [3, \infty), \text{ Axis of Symmetry: } x = -2, \text{ Vertex: } (-2, 3), \text{ Y-intercept: } (0, 7), \text{ Zero(s): None} \\
\text{2. Domain: } (-\infty, \infty), \text{ Range: } \left(-\infty, \frac{29}{4}\right], \text{ Axis of Symmetry: } x = -\frac{5}{2}, \text{ Vertex: } \left(-\frac{5}{2}, \frac{29}{4}\right), \text{ Y-intercept: } (0, 1), \text{ Zero(s): } x \approx -5.2 \text{ and } x \approx 0.2 \\
\text{3. Graph with Vertex: } \left(-\frac{1}{4}, -\frac{49}{8}\right), \text{ Y-intercept: } (0, -6), \text{ Zeros: } x = \frac{3}{2} \text{ and } x = -2
\end{array}
}
\]