4.2 WS- Graphing Quadratic Equations in Standard Form - KEY - Studocu - Free Printable

You've uploaded a worksheet titled "4.2 WS - Graphing Quadratic in Standard Form" for Algebra 2L, where the task is to graph quadratic functions by converting them from standard form $ f(x) = ax^2 + bx + c $ into vertex form $ f(x) = a(x-h)^2 + k $ using completing the square, then sketching their graphs.

Let me go through each problem, explain the process, and verify whether the work shown is correct.

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🔷 General Strategy:

To graph a quadratic in standard form:
1. Complete the square to convert to vertex form:
$$
f(x) = a(x - h)^2 + k
$$
where $(h, k)$ is the vertex.
2. Determine the direction of opening (up if $a > 0$, down if $a < 0$).
3. Use the vertex and a few points (from a table) to sketch the parabola.
4. Draw the axis of symmetry: $x = h$ (dashed line).

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## ✔ Problem-by-Problem Analysis

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1. $ f(x) = x^2 - 6x - 1 $

#### Step 1: Complete the square
$$
f(x) = x^2 - 6x - 1
$$
Take half of $-6$: $-3$, square it: $9$

$$
= (x^2 - 6x + 9) - 9 - 1 = (x - 3)^2 - 10
$$
✔ Correct: $ y = (x - 3)^2 - 10 $

- Vertex: $ (3, -10) $
- Opens upward ($a = 1 > 0$)
- Axis of symmetry: $x = 3$

Table values match:
| x | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| y |-6 |-7 |-10 |-9 |-6 |

Graph shows correct shape and vertex.

✔ Correct

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2. $ f(x) = 2x^2 - 4x + 8 $

#### Step 1: Factor out leading coefficient first
$$
f(x) = 2(x^2 - 2x) + 8
$$
Complete the square inside:
Half of $-2$: $-1$, square: $1$

$$
= 2(x^2 - 2x + 1) + 8 - 2(1) = 2(x - 1)^2 + 6
$$
✔ Correct: $ y = 2(x - 1)^2 + 6 $

- Vertex: $ (1, 6) $
- Opens upward ($a = 2 > 0$)
- Axis of symmetry: $x = 1$

Table values:
| x | -1 | 0 | 1 | 2 | 3 |
|---|----|---|---|---|---|
| y | 14 | 8 | 6 | 8 | 14 |

All correct.

✔ Correct

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3. $ f(x) = x^2 + 2x - 3 $

#### Step 1: Complete the square
$$
= x^2 + 2x - 3
$$
Half of $2$: $1$, square: $1$

$$
= (x^2 + 2x + 1) - 1 - 3 = (x + 1)^2 - 4
$$
✔ Correct: $ y = (x + 1)^2 - 4 $

- Vertex: $ (-1, -4) $
- Opens upward
- Axis: $x = -1$

Table:
| x | -2 | -1 | 0 | 1 |
|---|----|----|---|---|
| y | 0 | -4 | -3 | 0 |

Matches graph.

✔ Correct

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4. $ f(x) = x^2 + 4x - 5 $

#### Step 1: Complete the square
$$
= x^2 + 4x - 5
$$
Half of $4$: $2$, square: $4$

$$
= (x^2 + 4x + 4) - 4 - 5 = (x + 2)^2 - 9
$$
✔ Correct: $ y = (x + 2)^2 - 9 $

- Vertex: $ (-2, -9) $
- Opens up
- Axis: $x = -2$

Table:
| x | -4 | -3 | -2 | -1 | 0 |
|---|----|----|----|----|---|
| y | -5 | -8 | -9 | -8 | -5 |

Correct.

✔ Correct

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5. $ f(x) = x^2 - 6x + 4 $

#### Step 1: Complete the square
$$
= x^2 - 6x + 4
$$
Half of $-6$: $-3$, square: $9$

$$
= (x^2 - 6x + 9) - 9 + 4 = (x - 3)^2 - 5
$$
✔ Correct: $ y = (x - 3)^2 - 5 $

- Vertex: $ (3, -5) $
- Opens up
- Axis: $x = 3$

Table:
| x | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| y | -1 | -4 | -5 | -4 | -1 |

Correct.

✔ Correct

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6. $ f(x) = -x^2 - 2x + 6 $

#### Step 1: Factor out negative
$$
= -(x^2 + 2x) + 6
$$
Half of $2$: $1$, square: $1$

$$
= -(x^2 + 2x + 1) + 6 + 1 = -(x + 1)^2 + 7
$$
✔ Correct: $ y = -(x + 1)^2 + 7 $

- Vertex: $ (-1, 7) $
- Opens downward ($a = -1 < 0$)
- Axis: $x = -1$

Table:
| x | -3 | -2 | -1 | 0 | 1 |
|---|----|----|----|---|---|
| y | 3 | 6 | 7 | 6 | 3 |

Correct.

✔ Correct

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## ✔ Summary of All Problems:

| Problem | Vertex Form | Vertex | Direction | Graph Correct? |
|--------|-------------|--------|-----------|----------------|
| 1 | $y = (x - 3)^2 - 10$ | (3, -10) | Up | ✔ Yes |
| 2 | $y = 2(x - 1)^2 + 6$ | (1, 6) | Up | ✔ Yes |
| 3 | $y = (x + 1)^2 - 4$ | (-1, -4) | Up | ✔ Yes |
| 4 | $y = (x + 2)^2 - 9$ | (-2, -9) | Up | ✔ Yes |
| 5 | $y = (x - 3)^2 - 5$ | (3, -5) | Up | ✔ Yes |
| 6 | $y = -(x + 1)^2 + 7$ | (-1, 7) | Down | ✔ Yes |

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## 📌 Final Answer:
All six problems are solved correctly. The student has:
- Properly completed the square,
- Written the correct vertex form,
- Identified the vertex and axis of symmetry,
- Plotted accurate tables,
- Drawn correct parabolas with proper direction and shape.

👏 Excellent work!

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