Transformations of quadratic equations(vertex form) worksheet ... - Free Printable

Problem Analysis and Solution

The provided worksheet involves two main tasks:
1. Converting quadratic equations to vertex form.
2. Graphing a quadratic function in vertex form.

Let's solve each part step by step.

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#### Part 1: Choosing the Vertex Form for Each Equation

The general form of a quadratic equation is:
\[ y = ax^2 + bx + c \]

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

To convert from standard form to vertex form, we use the method of completing the square.

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##### Question 8: \( y = x^2 + 9x + 8 \)

1. Start with the given equation:
\[ y = x^2 + 9x + 8 \]

2. Focus on the \(x^2 + 9x\) part. To complete the square:
- Take the coefficient of \(x\), which is 9.
- Divide it by 2: \( \frac{9}{2} \).
- Square it: \( \left( \frac{9}{2} \right)^2 = \frac{81}{4} \).

3. Add and subtract \( \frac{81}{4} \) inside the equation:
\[
y = x^2 + 9x + \frac{81}{4} - \frac{81}{4} + 8
\]

4. Rewrite the perfect square trinomial:
\[
y = \left( x + \frac{9}{2} \right)^2 - \frac{81}{4} + 8
\]

5. Simplify the constant terms:
\[
-\frac{81}{4} + 8 = -\frac{81}{4} + \frac{32}{4} = -\frac{49}{4}
\]

6. The vertex form is:
\[
y = \left( x + \frac{9}{2} \right)^2 - \frac{49}{4}
\]

Answer: A

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##### Question 9: \( y = x^2 - 6x + 3 \)

1. Start with the given equation:
\[ y = x^2 - 6x + 3 \]

2. Focus on the \(x^2 - 6x\) part. To complete the square:
- Take the coefficient of \(x\), which is -6.
- Divide it by 2: \( \frac{-6}{2} = -3 \).
- Square it: \( (-3)^2 = 9 \).

3. Add and subtract 9 inside the equation:
\[
y = x^2 - 6x + 9 - 9 + 3
\]

4. Rewrite the perfect square trinomial:
\[
y = (x - 3)^2 - 9 + 3
\]

5. Simplify the constant terms:
\[
-9 + 3 = -6
\]

6. The vertex form is:
\[
y = (x - 3)^2 - 6
\]

Answer: B

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##### Question 10: \( y = -2x^2 + 5x \)

1. Start with the given equation:
\[ y = -2x^2 + 5x \]

2. Factor out the coefficient of \(x^2\) (which is -2) from the first two terms:
\[
y = -2 \left( x^2 - \frac{5}{2}x \right)
\]

3. Focus on the \(x^2 - \frac{5}{2}x\) part. To complete the square:
- Take the coefficient of \(x\), which is \(-\frac{5}{2}\).
- Divide it by 2: \( \frac{-\frac{5}{2}}{2} = -\frac{5}{4} \).
- Square it: \( \left( -\frac{5}{4} \right)^2 = \frac{25}{16} \).

4. Add and subtract \( \frac{25}{16} \) inside the parentheses:
\[
y = -2 \left( x^2 - \frac{5}{2}x + \frac{25}{16} - \frac{25}{16} \right)
\]

5. Rewrite the perfect square trinomial:
\[
y = -2 \left( \left( x - \frac{5}{4} \right)^2 - \frac{25}{16} \right)
\]

6. Distribute the -2:
\[
y = -2 \left( x - \frac{5}{4} \right)^2 + \frac{50}{16}
\]

7. Simplify the constant term:
\[
\frac{50}{16} = \frac{25}{8}
\]

8. The vertex form is:
\[
y = -2 \left( x - \frac{5}{4} \right)^2 + \frac{25}{8}
\]

Answer: A

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##### Question 17: \( y = -4x^2 - 24x - 15 \)

1. Start with the given equation:
\[ y = -4x^2 - 24x - 15 \]

2. Factor out the coefficient of \(x^2\) (which is -4) from the first two terms:
\[
y = -4 \left( x^2 + 6x \right) - 15
\]

3. Focus on the \(x^2 + 6x\) part. To complete the square:
- Take the coefficient of \(x\), which is 6.
- Divide it by 2: \( \frac{6}{2} = 3 \).
- Square it: \( 3^2 = 9 \).

4. Add and subtract 9 inside the parentheses:
\[
y = -4 \left( x^2 + 6x + 9 - 9 \right) - 15
\]

5. Rewrite the perfect square trinomial:
\[
y = -4 \left( (x + 3)^2 - 9 \right) - 15
\]

6. Distribute the -4:
\[
y = -4(x + 3)^2 + 36 - 15
\]

7. Simplify the constant terms:
\[
36 - 15 = 21
\]

8. The vertex form is:
\[
y = -4(x + 3)^2 + 21
\]

Answer: B

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#### Part 2: Graphing the Quadratic Function

##### Question 20: FIREWORKS

The height \( h \) of the rocket after \( t \) seconds is given by:
\[ h = -4.9(t - 4)^2 + 80 \]

This is already in vertex form:
\[ h = a(t - h)^2 + k \]
where:
- \( a = -4.9 \)
- \( h = 4 \)
- \( k = 80 \)

The vertex of the parabola is at \( (4, 80) \). Since \( a = -4.9 \) is negative, the parabola opens downward.

1. Vertex: The highest point of the rocket is at \( (4, 80) \).
2. Axis of Symmetry: The axis of symmetry is \( t = 4 \).
3. Shape: The parabola opens downward because \( a < 0 \).

Now, let's match this to the given graphs:
- Graph A: Opens upward (incorrect).
- Graph B: Opens downward but does not have the correct vertex or shape (incorrect).
- Graph C: Opens downward, has the vertex at \( (4, 80) \), and matches the description (correct).

Answer: C

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Final Answers:

1. Question 8: A
2. Question 9: B
3. Question 10: A
4. Question 17: B
5. Question 20: C

\[
\boxed{\text{A, B, A, B, C}}
\]