Worksheet for graphing parabolas equations with four examples.

It looks like you've uploaded a worksheet titled "Graphing Parabolas Equations" with four problems, each involving graphing a parabola. The graphs are already drawn in red on the coordinate grids. I'll go through each problem step by step to verify and explain the solution.

---

🔹 Problem 1:

Equation: $ y = -2x^2 + 12x - 16 $

#### Step-by-step Analysis:
This is a quadratic equation in standard form: $ y = ax^2 + bx + c $, where:
- $ a = -2 $
- $ b = 12 $
- $ c = -16 $

Since $ a < 0 $, the parabola opens downward.

#### Vertex:
Use vertex formula:
$ x = -\frac{b}{2a} = -\frac{12}{2(-2)} = \frac{12}{4} = 3 $

Now plug $ x = 3 $ into the equation:
$$
y = -2(3)^2 + 12(3) - 16 = -2(9) + 36 - 16 = -18 + 36 - 16 = 2
$$
So, the vertex is at (3, 2).

#### Y-intercept:
Set $ x = 0 $:
$$
y = -2(0)^2 + 12(0) - 16 = -16
$$
So, y-intercept is $ (0, -16) $

#### X-intercepts:
Solve $ -2x^2 + 12x - 16 = 0 $

Divide entire equation by -2:
$$
x^2 - 6x + 8 = 0
$$
Factor:
$$
(x - 4)(x - 2) = 0 \Rightarrow x = 2, 4
$$

So, x-intercepts: $ (2, 0) $ and $ (4, 0) $

✔ The graph shown has:
- Opens downward ✔
- Vertex around (3, 2) ✔
- Passes through (2,0), (4,0), (0,-16) ✔

✔️ Correctly graphed.

---

🔹 Problem 2:

Equation: $ y = -x^2 - 2x - 4 $

Standard form: $ y = ax^2 + bx + c $
- $ a = -1 $, so it opens downward
- $ b = -2 $, $ c = -4 $

#### Vertex:
$ x = -\frac{b}{2a} = -\frac{-2}{2(-1)} = \frac{2}{-2} = -1 $

Plug $ x = -1 $ into equation:
$$
y = -(-1)^2 - 2(-1) - 4 = -1 + 2 - 4 = -3
$$
Vertex: $ (-1, -3) $

#### Y-intercept:
$ x = 0 \Rightarrow y = -0 - 0 - 4 = -4 $ → $ (0, -4) $

#### X-intercepts:
Solve $ -x^2 - 2x - 4 = 0 $

Multiply both sides by -1:
$ x^2 + 2x + 4 = 0 $

Discriminant: $ D = 2^2 - 4(1)(4) = 4 - 16 = -12 < 0 $

No real solutions → no x-intercepts

So, the parabola does not cross the x-axis.

✔ Graph shows:
- Opens downward ✔
- Vertex at (-1, -3) ✔
- No x-intercepts ✔
- Passes through (0, -4) ✔

✔️ Correctly graphed.

---

🔹 Problem 3:

Equation: $ x = (y - 2)^2 - 3 $

This is a horizontal parabola (opens left or right).

Standard form: $ x = a(y - k)^2 + h $

Here:
- $ a = 1 $ → positive → opens to the right
- Vertex: $ (h, k) = (-3, 2) $

So, vertex is at $ (-3, 2) $

Let’s find a few points:

| y | x = (y - 2)^2 - 3 |
|---|---------------------|
| 0 | (0 - 2)^2 - 3 = 4 - 3 = 1 → (1, 0) |
| 1 | (1 - 2)^2 - 3 = 1 - 3 = -2 → (-2, 1) |
| 3 | (3 - 2)^2 - 3 = 1 - 3 = -2 → (-2, 3) |
| 4 | (4 - 2)^2 - 3 = 4 - 3 = 1 → (1, 4) |

So, symmetric about $ y = 2 $, opens to the right.

✔ Graph shows:
- Opens to the right ✔
- Vertex at (-3, 2) ✔
- Points like (1,0), (1,4), (-2,1), (-2,3) ✔

✔️ Correctly graphed.

---

🔹 Problem 4:

Equation: $ x = -2(y + 3)^2 + 1 $

Again, horizontal parabola.

Form: $ x = a(y - k)^2 + h $

Rewrite: $ x = -2(y - (-3))^2 + 1 $

So:
- $ a = -2 $ → negative → opens to the left
- Vertex: $ (1, -3) $

Let’s compute a few points:

| y | x = -2(y + 3)^2 + 1 |
|---|------------------------|
| -5 | -2(-5+3)^2 + 1 = -2(4) + 1 = -8 + 1 = -7 → (-7, -5) |
| -4 | -2(-4+3)^2 + 1 = -2(1) + 1 = -2 + 1 = -1 → (-1, -4) |
| -3 | -2(0)^2 + 1 = 1 → (1, -3) ← vertex |
| -2 | -2(-2+3)^2 + 1 = -2(1) + 1 = -1 → (-1, -2) |
| -1 | -2(-1+3)^2 + 1 = -2(4) + 1 = -8 + 1 = -7 → (-7, -1) |

So, symmetric about $ y = -3 $, opens to the left.

✔ Graph shows:
- Opens to the left ✔
- Vertex at (1, -3) ✔
- Points like (-1, -4), (-1, -2), (-7, -5), (-7, -1) ✔

✔️ Correctly graphed.

---

✔ Final Summary:

All four graphs are correctly drawn based on their equations.

| Problem | Equation | Direction | Vertex | Intercepts / Features | Correct? |
|--------|---------|-----------|--------|------------------------|----------|
| 1 | $ y = -2x^2 + 12x - 16 $ | Down | (3, 2) | x=2,4; y=-16 | ✔️ Yes |
| 2 | $ y = -x^2 - 2x - 4 $ | Down | (-1, -3) | No x-ints; y=-4 | ✔️ Yes |
| 3 | $ x = (y - 2)^2 - 3 $ | Right | (-3, 2) | Opens right | ✔️ Yes |
| 4 | $ x = -2(y + 3)^2 + 1 $ | Left | (1, -3) | Opens left | ✔️ Yes |

---

📌 Key Concepts Used:

- Vertex formula: $ x = -\frac{b}{2a} $ for vertical parabolas
- Standard forms:
- Vertical: $ y = a(x-h)^2 + k $
- Horizontal: $ x = a(y-k)^2 + h $
- Direction of opening depends on sign of coefficient
- Intercepts found by setting variables to zero
- Symmetry about vertex axis

Let me know if you'd like a printable version or need help generating similar problems!