Standard To Vertex Form - Fill Online, Printable, Fillable, Blank ... - Free Printable

Problem Overview:

The task involves working with quadratic equations in Vertex Form and converting them to Standard Form. Additionally, you need to identify the vertex of each parabola and determine whether it opens up or down. Finally, for each problem, you are asked to write the equation of the parabola after it has been translated based on given instructions.

Given Information:

1. Vertex Form: The equations are initially provided in Vertex Form.
2. Standard Form: You need to convert these equations to Standard Form.
3. Vertex Identification: Identify the vertex of each parabola.
4. Direction of Opening: Determine whether the parabola opens up or down.
5. Translation Instructions: Apply the given translation instructions to modify the original equation.

Step-by-Step Solution:

#### 1. Equation 1: \( y = (x - 3)^2 - 10 \)

- Vertex Form: \( y = (x - 3)^2 - 10 \)
- Vertex: The vertex form is \( y = a(x - h)^2 + k \), where \( (h, k) \) is the vertex. Here, \( h = 3 \) and \( k = -10 \). So, the vertex is \( (3, -10) \).
- Direction of Opening: Since the coefficient of \( (x - 3)^2 \) is positive (\( a = 1 \)), the parabola opens up.
- Standard Form: Expand the equation:
\[
y = (x - 3)^2 - 10
\]
\[
y = x^2 - 6x + 9 - 10
\]
\[
y = x^2 - 6x - 1
\]
So, the Standard Form is \( y = x^2 - 6x - 1 \).

- Translation Instruction: "6 left, 7 down"
- Translating 6 units left means replacing \( x \) with \( x + 6 \).
- Translating 7 units down means subtracting 7 from the entire equation.
- New equation:
\[
y = ((x + 6) - 3)^2 - 10 - 7
\]
\[
y = (x + 3)^2 - 17
\]

#### 2. Equation 2: \( y = (x + 5)^2 + 4 \)

- Vertex Form: \( y = (x + 5)^2 + 4 \)
- Vertex: Here, \( h = -5 \) and \( k = 4 \). So, the vertex is \( (-5, 4) \).
- Direction of Opening: Since the coefficient of \( (x + 5)^2 \) is positive (\( a = 1 \)), the parabola opens up.
- Standard Form: Expand the equation:
\[
y = (x + 5)^2 + 4
\]
\[
y = x^2 + 10x + 25 + 4
\]
\[
y = x^2 + 10x + 29
\]
So, the Standard Form is \( y = x^2 + 10x + 29 \).

- Translation Instruction: "8 left, 4 up"
- Translating 8 units left means replacing \( x \) with \( x + 8 \).
- Translating 4 units up means adding 4 to the entire equation.
- New equation:
\[
y = ((x + 8) + 5)^2 + 4 + 4
\]
\[
y = (x + 13)^2 + 8
\]

#### 3. Equation 3: \( y = 3(x - 2)^2 - 6 \)

- Vertex Form: \( y = 3(x - 2)^2 - 6 \)
- Vertex: Here, \( h = 2 \) and \( k = -6 \). So, the vertex is \( (2, -6) \).
- Direction of Opening: Since the coefficient of \( (x - 2)^2 \) is positive (\( a = 3 \)), the parabola opens up.
- Standard Form: Expand the equation:
\[
y = 3(x - 2)^2 - 6
\]
\[
y = 3(x^2 - 4x + 4) - 6
\]
\[
y = 3x^2 - 12x + 12 - 6
\]
\[
y = 3x^2 - 12x + 6
\]
So, the Standard Form is \( y = 3x^2 - 12x + 6 \).

- Translation Instruction: "3 right, 2 down"
- Translating 3 units right means replacing \( x \) with \( x - 3 \).
- Translating 2 units down means subtracting 2 from the entire equation.
- New equation:
\[
y = 3((x - 3) - 2)^2 - 6 - 2
\]
\[
y = 3(x - 5)^2 - 8
\]

#### 4. Equation 4: \( y = -4x^2 - 24x - 32 \)

- Vertex Form: First, we need to convert this to Vertex Form. The given equation is already in Standard Form: \( y = -4x^2 - 24x - 32 \).
- Vertex: To find the vertex, use the formula \( x = -\frac{b}{2a} \):
\[
a = -4, \quad b = -24
\]
\[
x = -\frac{-24}{2(-4)} = -\frac{-24}{-8} = -3
\]
Substitute \( x = -3 \) back into the equation to find \( y \):
\[
y = -4(-3)^2 - 24(-3) - 32
\]
\[
y = -4(9) + 72 - 32
\]
\[
y = -36 + 72 - 32
\]
\[
y = 4
\]
So, the vertex is \( (-3, 4) \).
- Direction of Opening: Since the coefficient of \( x^2 \) is negative (\( a = -4 \)), the parabola opens down.
- Vertex Form: Rewrite the equation in Vertex Form by completing the square:
\[
y = -4(x^2 + 6x) - 32
\]
\[
y = -4(x^2 + 6x + 9 - 9) - 32
\]
\[
y = -4((x + 3)^2 - 9) - 32
\]
\[
y = -4(x + 3)^2 + 36 - 32
\]
\[
y = -4(x + 3)^2 + 4
\]
So, the Vertex Form is \( y = -4(x + 3)^2 + 4 \).

- Translation Instruction: "9 right, 4 up"
- Translating 9 units right means replacing \( x \) with \( x - 9 \).
- Translating 4 units up means adding 4 to the entire equation.
- New equation:
\[
y = -4((x - 9) + 3)^2 + 4 + 4
\]
\[
y = -4(x - 6)^2 + 8
\]

Final Answer:

\[
\boxed{
\begin{array}{|c|c|c|c|}
\hline
\text{Original Equation} & \text{Original Vertex} & \text{Translation} & \text{New Equation} \\
\hline
y = (x - 3)^2 - 10 & (3, -10) & 6 \text{ left}, 7 \text{ down} & y = (x + 3)^2 - 17 \\
\hline
y = (x + 5)^2 + 4 & (-5, 4) & 8 \text{ left}, 4 \text{ up} & y = (x + 13)^2 + 8 \\
\hline
y = 3(x - 2)^2 - 6 & (2, -6) & 3 \text{ right}, 2 \text{ down} & y = 3(x - 5)^2 - 8 \\
\hline
y = -4x^2 - 24x - 32 & (-3, 4) & 9 \text{ right}, 4 \text{ up} & y = -4(x - 6)^2 + 8 \\
\hline
\end{array}
}
\]