Graphing Quadratics Notes and Worksheets - Lindsay Bowden - Free Printable

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:
- Function: \( f(x) = x^2 + 4x + 7 \)
- Graph is provided.

##### Task:
- Find the domain, range, axis of symmetry, vertex, y-intercept, and zeros.

##### Solution:

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

2. Range:
- The graph shows a parabola opening upwards (since the coefficient of \( x^2 \) is positive).
- The vertex is the minimum point of the parabola.
- From the graph, the vertex appears to be at \( (x, y) = (-2, 3) \).
- Therefore, the range is all \( y \)-values greater than or equal to the \( y \)-coordinate of the vertex.
- Range: \( [3, \infty) \)

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 \( b = 4 \):
\[
x = -\frac{4}{2 \cdot 1} = -2
\]
- Axis of Symmetry: \( x = -2 \)

4. Vertex:
- The vertex is the point where the axis of symmetry intersects the parabola.
- From the graph, the vertex is at \( (-2, 3) \).
- Vertex: \( (-2, 3) \)

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

6. Zeros:
- The zeros are the \( x \)-values where the graph intersects the \( x \)-axis (i.e., \( f(x) = 0 \)).
- From the graph, the parabola does not intersect the \( x \)-axis.
- Zeros: None

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

##### Given:
- Function: \( y = -x^2 - 5x + 1 \)
- Graph is provided.

##### Task:
- Find the domain, range, axis of symmetry, vertex, y-intercept, and zeros.

##### Solution:

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

2. Range:
- The graph shows a parabola opening downwards (since the coefficient of \( x^2 \) is negative).
- The vertex is the maximum point of the parabola.
- From the graph, the vertex appears to be at \( (x, y) = \left(-\frac{5}{2}, \frac{29}{4}\right) \).
- Therefore, the range is all \( y \)-values less than or equal to the \( y \)-coordinate of the vertex.
- Range: \( \left(-\infty, \frac{29}{4}\right] \)

3. Axis of Symmetry:
- The axis of symmetry is given by \( x = -\frac{b}{2a} \).
- Here, \( a = -1 \) and \( b = -5 \):
\[
x = -\frac{-5}{2(-1)} = -\frac{5}{2}
\]
- Axis of Symmetry: \( x = -\frac{5}{2} \)

4. Vertex:
- The vertex is the point where the axis of symmetry intersects the parabola.
- From the graph, the vertex is at \( \left(-\frac{5}{2}, \frac{29}{4}\right) \).
- Vertex: \( \left(-\frac{5}{2}, \frac{29}{4}\right) \)

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

6. Zeros:
- The zeros are the \( x \)-values where the graph intersects the \( x \)-axis (i.e., \( y = 0 \)).
- From the graph, the parabola intersects the \( x \)-axis at two points.
- Approximate values from the graph: \( x \approx -5.2 \) and \( x \approx 0.2 \).
- Zeros: \( x \approx -5.2 \) and \( x \approx 0.2 \)

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

##### Given:
- Function: \( f(x) = 2x^2 + x - 6 \)

##### Task:
- Graph the quadratic function.

##### Solution:

1. Identify Key Features:
- Vertex: The vertex is found using \( x = -\frac{b}{2a} \).
- Here, \( a = 2 \) and \( b = 1 \):
\[
x = -\frac{1}{2 \cdot 2} = -\frac{1}{4}
\]
- Substitute \( x = -\frac{1}{4} \) into the function to find the \( y \)-coordinate:
\[
f\left(-\frac{1}{4}\right) = 2\left(-\frac{1}{4}\right)^2 + \left(-\frac{1}{4}\right) - 6 = 2\left(\frac{1}{16}\right) - \frac{1}{4} - 6 = \frac{2}{16} - \frac{4}{16} - \frac{96}{16} = -\frac{98}{16} = -\frac{49}{8}
\]
- Vertex: \( \left(-\frac{1}{4}, -\frac{49}{8}\right) \)

- Y-intercept: Set \( x = 0 \):
\[
f(0) = 2(0)^2 + (0) - 6 = -6
\]
- Y-intercept: \( (0, -6) \)

- 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 \( c = -6 \):
\[
x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-6)}}{2(2)} = \frac{-1 \pm \sqrt{1 + 48}}{4} = \frac{-1 \pm \sqrt{49}}{4} = \frac{-1 \pm 7}{4}
\]
\[
x = \frac{6}{4} = \frac{3}{2} \quad \text{or} \quad x = \frac{-8}{4} = -2
\]
- Zeros: \( x = \frac{3}{2} \) and \( x = -2 \)

2. Plot Points:
- Use the vertex, y-intercept, and zeros to plot the graph.
- Additional points can be calculated by substituting other \( x \)-values into the function.

3. Draw the Parabola:
- Since \( a = 2 \) is positive, the parabola opens upwards.
- Sketch the parabola through the plotted points.

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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: \( (0, 7) \)
- Zeros: 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: \( (0, 1) \)
- Zeros: \( x \approx -5.2 \) and \( x \approx 0.2 \)

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

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