Algebra 2 Worksheets | Quadratic Functions and Inequalities Worksheets - Free Printable

Let's solve each of the six parabola problems step by step, identifying the requested properties:

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General Concepts:

For a parabola, we analyze its equation to determine:
- Vertex: The turning point.
- Axis of Symmetry: A vertical or horizontal line passing through the vertex.
- Opens: Direction (up/down for vertical; left/right for horizontal).
- Min/Max value: The y-value (if vertical) or x-value (if horizontal) at the vertex.
- Latus rectum (Latus): Length of the latus rectum = $ |4p| $, where $ p $ is the distance from vertex to focus.
- Intercepts: Where the graph crosses axes.

We’ll go through each problem.

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## 1) $ x = -2y^2 - 4y + 70 $

This is a horizontal parabola (x in terms of y).

Step 1: Rewrite in vertex form

We complete the square.

$$
x = -2(y^2 + 2y) + 70
$$

Complete the square inside the parentheses:
$$
y^2 + 2y \rightarrow (y+1)^2 - 1
$$

So:
$$
x = -2[(y+1)^2 - 1] + 70 = -2(y+1)^2 + 2 + 70 = -2(y+1)^2 + 72
$$

Vertex form: $ x = -2(y + 1)^2 + 72 $

Now extract properties:

- Vertex: $ (72, -1) $
- Axis of Symmetry: $ y = -1 $
- Opens: Left (because coefficient of $ (y+1)^2 $ is negative)
- Min/Max value: Since it opens left, maximum x-value occurs at vertex → Max x = 72
- Latus: For $ x = a(y-k)^2 + h $, the latus rectum length is $ \frac{1}{|a|} \times 4 $, but wait — actually, standard form:
$ x = \frac{1}{4p}(y-k)^2 + h $. So here:
$ x = -2(y+1)^2 + 72 $ → compare with $ x = \frac{1}{4p}(y-k)^2 + h $
So $ \frac{1}{4p} = -2 $ → $ 4p = -\frac{1}{2} $ → $ |4p| = \frac{1}{2} $
So Latus = $ \frac{1}{2} $

- y-intercept: Set $ x = 0 $:
$$
0 = -2y^2 - 4y + 70 \Rightarrow 2y^2 + 4y - 70 = 0 \Rightarrow y^2 + 2y - 35 = 0
$$
$$
y = \frac{-2 \pm \sqrt{4 + 140}}{2} = \frac{-2 \pm \sqrt{144}}{2} = \frac{-2 \pm 12}{2}
\Rightarrow y = 5, -7
$$
So y-int: (0, 5), (0, -7)

- x-intercept: Set $ y = 0 $:
$$
x = -2(0)^2 - 4(0) + 70 = 70
$$
So x-int: (70, 0)

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✔ Summary for #1:

- Min/Max value: Max x = 72
- Latus: $ \frac{1}{2} $
- y-int: (0, 5), (0, -7)
- x-int: (70, 0)
- Vertex: (72, -1)
- Axis of Symmetry: $ y = -1 $
- Opens: Left

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## 2) $ y = 2x^2 - 4x - 6 $

Vertical parabola.

Complete the square:

$$
y = 2(x^2 - 2x) - 6
$$

Complete the square:
$$
x^2 - 2x \rightarrow (x-1)^2 - 1
$$

So:
$$
y = 2[(x-1)^2 - 1] - 6 = 2(x-1)^2 - 2 - 6 = 2(x-1)^2 - 8
$$

Vertex form: $ y = 2(x - 1)^2 - 8 $

Properties:

- Vertex: $ (1, -8) $
- Axis of Symmetry: $ x = 1 $
- Opens: Up (coefficient positive)
- Min/Max value: Minimum y-value = -8
- Latus: Use $ y = \frac{1}{4p}(x-h)^2 + k $. Here, $ \frac{1}{4p} = 2 $ → $ 4p = \frac{1}{2} $ → Latus = $ \frac{1}{2} $
- y-intercept: Set $ x = 0 $:
$$
y = 2(0)^2 - 4(0) - 6 = -6 \Rightarrow (0, -6)
$$
- x-intercepts: Set $ y = 0 $:
$$
2x^2 - 4x - 6 = 0 \Rightarrow x^2 - 2x - 3 = 0 \Rightarrow (x-3)(x+1) = 0
\Rightarrow x = 3, -1
$$
So x-int: (-1, 0), (3, 0)

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✔ Summary for #2:

- Min/Max value: Min y = -8
- Latus: $ \frac{1}{2} $
- y-int: (0, -6)
- x-int: (-1, 0), (3, 0)
- Vertex: (1, -8)
- Axis of Symmetry: $ x = 1 $
- Opens: Up

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## 3) $ y = -2(x - 3)^2 + 2 $

Already in vertex form!

- Vertex: $ (3, 2) $
- Axis of Symmetry: $ x = 3 $
- Opens: Down (coefficient negative)
- Min/Max value: Maximum y = 2
- Latus: Coefficient is -2 → $ \frac{1}{4p} = -2 $ → $ 4p = -\frac{1}{2} $ → $ |4p| = \frac{1}{2} $ → Latus = $ \frac{1}{2} $
- y-intercept: Set $ x = 0 $:
$$
y = -2(0 - 3)^2 + 2 = -2(9) + 2 = -18 + 2 = -16 \Rightarrow (0, -16)
$$
- x-intercepts: Set $ y = 0 $:
$$
0 = -2(x - 3)^2 + 2 \Rightarrow 2(x - 3)^2 = 2 \Rightarrow (x - 3)^2 = 1 \Rightarrow x = 4, 2
$$
So x-int: (2, 0), (4, 0)

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✔ Summary for #3:

- Min/Max value: Max y = 2
- Latus: $ \frac{1}{2} $
- y-int: (0, -16)
- x-int: (2, 0), (4, 0)
- Vertex: (3, 2)
- Axis of Symmetry: $ x = 3 $
- Opens: Down

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## 4) $ x = 2(y - 3)^2 - 2 $

Horizontal parabola, already in vertex form.

- Vertex: $ (-2, 3) $
- Axis of Symmetry: $ y = 3 $
- Opens: Right (coefficient positive)
- Min/Max value: Minimum x-value = -2
- Latus: $ \frac{1}{4p} = 2 $ → $ 4p = \frac{1}{2} $ → Latus = $ \frac{1}{2} $
- y-intercept: Set $ x = 0 $:
$$
0 = 2(y - 3)^2 - 2 \Rightarrow 2(y - 3)^2 = 2 \Rightarrow (y - 3)^2 = 1 \Rightarrow y = 4, 2
$$
So y-int: (0, 4), (0, 2)
- x-intercept: Set $ y = 0 $:
$$
x = 2(0 - 3)^2 - 2 = 2(9) - 2 = 18 - 2 = 16 \Rightarrow (16, 0)
$$

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✔ Summary for #4:

- Min/Max value: Min x = -2
- Latus: $ \frac{1}{2} $
- y-int: (0, 4), (0, 2)
- x-int: (16, 0)
- Vertex: (-2, 3)
- Axis of Symmetry: $ y = 3 $
- Opens: Right

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## 5) $ x = 2y^2 - 24y + 70 $

Horizontal parabola. Complete the square.

$$
x = 2(y^2 - 12y) + 70
$$

Complete the square:
$$
y^2 - 12y \rightarrow (y - 6)^2 - 36
$$

So:
$$
x = 2[(y - 6)^2 - 36] + 70 = 2(y - 6)^2 - 72 + 70 = 2(y - 6)^2 - 2
$$

Vertex form: $ x = 2(y - 6)^2 - 2 $

Properties:

- Vertex: $ (-2, 6) $
- Axis of Symmetry: $ y = 6 $
- Opens: Right (positive coefficient)
- Min/Max value: Minimum x = -2
- Latus: $ \frac{1}{4p} = 2 $ → $ 4p = \frac{1}{2} $ → Latus = $ \frac{1}{2} $
- y-intercept: Set $ x = 0 $:
$$
0 = 2y^2 - 24y + 70 \Rightarrow y^2 - 12y + 35 = 0 \Rightarrow (y - 5)(y - 7) = 0
\Rightarrow y = 5, 7
$$
So y-int: (0, 5), (0, 7)
- x-intercept: Set $ y = 0 $:
$$
x = 2(0)^2 - 24(0) + 70 = 70 \Rightarrow (70, 0)
$$

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✔ Summary for #5:

- Min/Max value: Min x = -2
- Latus: $ \frac{1}{2} $
- y-int: (0, 5), (0, 7)
- x-int: (70, 0)
- Vertex: (-2, 6)
- Axis of Symmetry: $ y = 6 $
- Opens: Right

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## 6) $ y = -2(x - 6)^2 + 2 $

Already in vertex form.

- Vertex: $ (6, 2) $
- Axis of Symmetry: $ x = 6 $
- Opens: Down (negative coefficient)
- Min/Max value: Maximum y = 2
- Latus: $ \frac{1}{4p} = -2 $ → $ 4p = -\frac{1}{2} $ → $ |4p| = \frac{1}{2} $ → Latus = $ \frac{1}{2} $
- y-intercept: Set $ x = 0 $:
$$
y = -2(0 - 6)^2 + 2 = -2(36) + 2 = -72 + 2 = -70 \Rightarrow (0, -70)
$$
- x-intercepts: Set $ y = 0 $:
$$
0 = -2(x - 6)^2 + 2 \Rightarrow 2(x - 6)^2 = 2 \Rightarrow (x - 6)^2 = 1 \Rightarrow x = 7, 5
$$
So x-int: (5, 0), (7, 0)

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✔ Summary for #6:

- Min/Max value: Max y = 2
- Latus: $ \frac{1}{2} $
- y-int: (0, -70)
- x-int: (5, 0), (7, 0)
- Vertex: (6, 2)
- Axis of Symmetry: $ x = 6 $
- Opens: Down

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Final Answer Table:

| Problem | Min/Max Value | Latus | y-int | x-int | Vertex | Axis of Symmetry | Opens |
|--------|----------------|-------|--------|--------|--------|--------------------|--------|
| 1 | Max x = 72 | ½ | (0,5),(0,-7) | (70,0) | (72,-1) | y = -1 | Left |
| 2 | Min y = -8 | ½ | (0,-6) | (-1,0),(3,0) | (1,-8) | x = 1 | Up |
| 3 | Max y = 2 | ½ | (0,-16) | (2,0),(4,0) | (3,2) | x = 3 | Down |
| 4 | Min x = -2 | ½ | (0,4),(0,2) | (16,0) | (-2,3) | y = 3 | Right |
| 5 | Min x = -2 | ½ | (0,5),(0,7) | (70,0) | (-2,6) | y = 6 | Right |
| 6 | Max y = 2 | ½ | (0,-70) | (5,0),(7,0) | (6,2) | x = 6 | Down |

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✔ All problems solved! Let me know if you'd like this formatted as a printable sheet.