Properties of Parabolas Worksheet for 9th - 11th Grade | Lesson Planet - Free Printable

Problem Analysis:

The task involves identifying the vertex of each given parabola and graphing them. The equations provided are in standard quadratic form, which can be rewritten into vertex form to easily identify the vertex. Let's solve this step by step.

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Part 1: Identify the Vertex of Each Parabola

#### Equation 1: \( y = x^2 - 6x - 7 \)

1. Rewrite in Vertex Form:
The general form of a quadratic equation is \( y = ax^2 + bx + c \). Here, \( a = 1 \), \( b = -6 \), and \( c = -7 \).

The vertex form of a parabola is \( y = a(x - h)^2 + k \), where \( (h, k) \) is the vertex.

To convert to vertex form, complete the square:
\[
y = x^2 - 6x - 7
\]
- Factor out the coefficient of \( x^2 \) (which is 1 in this case):
\[
y = (x^2 - 6x) - 7
\]
- Complete the square for \( x^2 - 6x \):
- Take half of the coefficient of \( x \) (which is \(-6\)), divide by 2 to get \(-3\), and square it to get \( 9 \).
- Add and subtract 9 inside the parentheses:
\[
y = (x^2 - 6x + 9 - 9) - 7
\]
\[
y = (x - 3)^2 - 9 - 7
\]
\[
y = (x - 3)^2 - 16
\]

- The equation is now in vertex form:
\[
y = (x - 3)^2 - 16
\]

2. Identify the Vertex:
From the vertex form \( y = a(x - h)^2 + k \), the vertex is \( (h, k) \). Here, \( h = 3 \) and \( k = -16 \).

Therefore, the vertex is:
\[
\boxed{(3, -16)}
\]

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#### Equation 2: \( y = x^2 + 10x + 25 \)

1. Rewrite in Vertex Form:
Here, \( a = 1 \), \( b = 10 \), and \( c = 25 \).

- Factor out the coefficient of \( x^2 \) (which is 1):
\[
y = (x^2 + 10x) + 25
\]
- Complete the square for \( x^2 + 10x \):
- Take half of the coefficient of \( x \) (which is \( 10 \)), divide by 2 to get \( 5 \), and square it to get \( 25 \).
- Add and subtract 25 inside the parentheses:
\[
y = (x^2 + 10x + 25 - 25) + 25
\]
\[
y = (x + 5)^2 - 25 + 25
\]
\[
y = (x + 5)^2
\]

- The equation is now in vertex form:
\[
y = (x + 5)^2
\]

2. Identify the Vertex:
From the vertex form \( y = a(x - h)^2 + k \), the vertex is \( (h, k) \). Here, \( h = -5 \) and \( k = 0 \).

Therefore, the vertex is:
\[
\boxed{(-5, 0)}
\]

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Part 2: Graph Each Equation

#### Graph 1: \( y = x^2 - 6x - 7 \)

1. Vertex: From Part 1, the vertex is \( (3, -16) \).
2. Axis of Symmetry: The axis of symmetry is \( x = h \), so \( x = 3 \).
3. Additional Points:
- Choose \( x \)-values around the vertex and calculate \( y \)-values:
- For \( x = 0 \):
\[
y = (0)^2 - 6(0) - 7 = -7 \quad \Rightarrow \quad (0, -7)
\]
- For \( x = 1 \):
\[
y = (1)^2 - 6(1) - 7 = 1 - 6 - 7 = -12 \quad \Rightarrow \quad (1, -12)
\]
- For \( x = 2 \):
\[
y = (2)^2 - 6(2) - 7 = 4 - 12 - 7 = -15 \quad \Rightarrow \quad (2, -15)
\]
- For \( x = 4 \):
\[
y = (4)^2 - 6(4) - 7 = 16 - 24 - 7 = -15 \quad \Rightarrow \quad (4, -15)
\]
- For \( x = 5 \):
\[
y = (5)^2 - 6(5) - 7 = 25 - 30 - 7 = -12 \quad \Rightarrow \quad (5, -12)
\]
- For \( x = 6 \):
\[
y = (6)^2 - 6(6) - 7 = 36 - 36 - 7 = -7 \quad \Rightarrow \quad (6, -7)
\]

Plot these points and draw a smooth parabola through them.

#### Graph 2: \( y = x^2 + 10x + 25 \)

1. Vertex: From Part 1, the vertex is \( (-5, 0) \).
2. Axis of Symmetry: The axis of symmetry is \( x = h \), so \( x = -5 \).
3. Additional Points:
- Choose \( x \)-values around the vertex and calculate \( y \)-values:
- For \( x = -8 \):
\[
y = (-8)^2 + 10(-8) + 25 = 64 - 80 + 25 = 9 \quad \Rightarrow \quad (-8, 9)
\]
- For \( x = -7 \):
\[
y = (-7)^2 + 10(-7) + 25 = 49 - 70 + 25 = 4 \quad \Rightarrow \quad (-7, 4)
\]
- For \( x = -6 \):
\[
y = (-6)^2 + 10(-6) + 25 = 36 - 60 + 25 = 1 \quad \Rightarrow \quad (-6, 1)
\]
- For \( x = -4 \):
\[
y = (-4)^2 + 10(-4) + 25 = 16 - 40 + 25 = 1 \quad \Rightarrow \quad (-4, 1)
\]
- For \( x = -3 \):
\[
y = (-3)^2 + 10(-3) + 25 = 9 - 30 + 25 = 4 \quad \Rightarrow \quad (-3, 4)
\]
- For \( x = -2 \):
\[
y = (-2)^2 + 10(-2) + 25 = 4 - 20 + 25 = 9 \quad \Rightarrow \quad (-2, 9)
\]

Plot these points and draw a smooth parabola through them.

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Part 3: Identify the Maximum or Minimum Value and Describe the Graph

#### Equation 3: \( f(x) = -x^2 + x + 20 \)

1. Rewrite in Vertex Form:
Here, \( a = -1 \), \( b = 1 \), and \( c = 20 \).

- Factor out the coefficient of \( x^2 \) (which is \(-1\)):
\[
f(x) = -(x^2 - x) + 20
\]
- Complete the square for \( x^2 - x \):
- Take half of the coefficient of \( x \) (which is \(-1\)), divide by 2 to get \(\frac{1}{2}\), and square it to get \(\frac{1}{4}\).
- Add and subtract \(\frac{1}{4}\) inside the parentheses:
\[
f(x) = -\left(x^2 - x + \frac{1}{4} - \frac{1}{4}\right) + 20
\]
\[
f(x) = -\left(\left(x - \frac{1}{2}\right)^2 - \frac{1}{4}\right) + 20
\]
\[
f(x) = -\left(x - \frac{1}{2}\right)^2 + \frac{1}{4} + 20
\]
\[
f(x) = -\left(x - \frac{1}{2}\right)^2 + \frac{81}{4}
\]

- The equation is now in vertex form:
\[
f(x) = -\left(x - \frac{1}{2}\right)^2 + \frac{81}{4}
\]

2. Identify the Vertex:
From the vertex form \( f(x) = a(x - h)^2 + k \), the vertex is \( (h, k) \). Here, \( h = \frac{1}{2} \) and \( k = \frac{81}{4} \).

Therefore, the vertex is:
\[
\left( \frac{1}{2}, \frac{81}{4} \right)
\]

3. Maximum or Minimum Value:
Since \( a = -1 \) (negative), the parabola opens downward, and the vertex represents the maximum value of the function.

The maximum value is:
\[
\boxed{\frac{81}{4}}
\]

4. Describe the Graph:
- The graph is a downward-opening parabola.
- The vertex is at \( \left( \frac{1}{2}, \frac{81}{4} \right) \).
- The axis of symmetry is \( x = \frac{1}{2} \).

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Final Answers:

1. Vertex of \( y = x^2 - 6x - 7 \): \(\boxed{(3, -16)}\)
2. Vertex of \( y = x^2 + 10x + 25 \): \(\boxed{(-5, 0)}\)
3. Maximum Value of \( f(x) = -x^2 + x + 20 \): \(\boxed{\frac{81}{4}}\)