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To find the equation of the quadratic function in both vertex and standard form given that the vertex is at [tex]\((-1, 6)\)[/tex] and the parabola passes through the point [tex]\((2, 24)\)[/tex], we can follow these steps:
### Step-by-Step Solution:
#### 1. Understanding the Vertex Form of a Quadratic Function:
The vertex form of a quadratic function is:
[tex]\[ y = a(x - h)^2 + k \][/tex]
where [tex]\((h, k)\)[/tex] is the vertex of the parabola. Here, the vertex [tex]\((h, k)\)[/tex] is given as [tex]\((-1, 6)\)[/tex], so [tex]\(h = -1\)[/tex] and [tex]\(k = 6\)[/tex].
#### 2. Plugging the Vertex into the Equation:
Substituting the vertex values into the vertex form equation, we get:
[tex]\[ y = a(x - (-1))^2 + 6 \][/tex]
[tex]\[ y = a(x + 1)^2 + 6 \][/tex]
#### 3. Using the Given Point to Find [tex]\(a\)[/tex]:
We know that the function passes through the point [tex]\((2, 24)\)[/tex]. Substituting [tex]\(x = 2\)[/tex] and [tex]\(y = 24\)[/tex] into the equation, we get:
[tex]\[ 24 = a(2 + 1)^2 + 6 \][/tex]
[tex]\[ 24 = a \cdot 3^2 + 6 \][/tex]
[tex]\[ 24 = 9a + 6 \][/tex]
Solving for [tex]\(a\)[/tex]:
[tex]\[ 24 - 6 = 9a \][/tex]
[tex]\[ 18 = 9a \][/tex]
[tex]\[ a = 2 \][/tex]
So, we have found that [tex]\(a = 2\)[/tex].
#### 4. Writing the Vertex Form:
Now that we have [tex]\(a = 2\)[/tex], [tex]\(h = -1\)[/tex], and [tex]\(k = 6\)[/tex], the vertex form of the quadratic function is:
[tex]\[ y = 2(x + 1)^2 + 6 \][/tex]
#### 5. Converting Vertex Form to Standard Form:
The standard form of a quadratic function is:
[tex]\[ y = ax^2 + bx + c \][/tex]
To convert [tex]\( y = 2(x + 1)^2 + 6 \)[/tex] to standard form, we need to expand it:
[tex]\[ y = 2(x + 1)^2 + 6 \][/tex]
[tex]\[ y = 2(x^2 + 2x + 1) + 6 \][/tex]
[tex]\[ y = 2x^2 + 4x + 2 + 6 \][/tex]
[tex]\[ y = 2x^2 + 4x + 8 \][/tex]
So, the quadratic function in standard form is:
[tex]\[ y = 2x^2 + 4x + 8 \][/tex]
#### Summary:
- Vertex Form: [tex]\( y = 2(x + 1)^2 + 6 \)[/tex]
- Standard Form: [tex]\( y = 2x^2 + 4x + 8 \)[/tex]
These are the equations of the quadratic function that satisfy the given conditions.
### Step-by-Step Solution:
#### 1. Understanding the Vertex Form of a Quadratic Function:
The vertex form of a quadratic function is:
[tex]\[ y = a(x - h)^2 + k \][/tex]
where [tex]\((h, k)\)[/tex] is the vertex of the parabola. Here, the vertex [tex]\((h, k)\)[/tex] is given as [tex]\((-1, 6)\)[/tex], so [tex]\(h = -1\)[/tex] and [tex]\(k = 6\)[/tex].
#### 2. Plugging the Vertex into the Equation:
Substituting the vertex values into the vertex form equation, we get:
[tex]\[ y = a(x - (-1))^2 + 6 \][/tex]
[tex]\[ y = a(x + 1)^2 + 6 \][/tex]
#### 3. Using the Given Point to Find [tex]\(a\)[/tex]:
We know that the function passes through the point [tex]\((2, 24)\)[/tex]. Substituting [tex]\(x = 2\)[/tex] and [tex]\(y = 24\)[/tex] into the equation, we get:
[tex]\[ 24 = a(2 + 1)^2 + 6 \][/tex]
[tex]\[ 24 = a \cdot 3^2 + 6 \][/tex]
[tex]\[ 24 = 9a + 6 \][/tex]
Solving for [tex]\(a\)[/tex]:
[tex]\[ 24 - 6 = 9a \][/tex]
[tex]\[ 18 = 9a \][/tex]
[tex]\[ a = 2 \][/tex]
So, we have found that [tex]\(a = 2\)[/tex].
#### 4. Writing the Vertex Form:
Now that we have [tex]\(a = 2\)[/tex], [tex]\(h = -1\)[/tex], and [tex]\(k = 6\)[/tex], the vertex form of the quadratic function is:
[tex]\[ y = 2(x + 1)^2 + 6 \][/tex]
#### 5. Converting Vertex Form to Standard Form:
The standard form of a quadratic function is:
[tex]\[ y = ax^2 + bx + c \][/tex]
To convert [tex]\( y = 2(x + 1)^2 + 6 \)[/tex] to standard form, we need to expand it:
[tex]\[ y = 2(x + 1)^2 + 6 \][/tex]
[tex]\[ y = 2(x^2 + 2x + 1) + 6 \][/tex]
[tex]\[ y = 2x^2 + 4x + 2 + 6 \][/tex]
[tex]\[ y = 2x^2 + 4x + 8 \][/tex]
So, the quadratic function in standard form is:
[tex]\[ y = 2x^2 + 4x + 8 \][/tex]
#### Summary:
- Vertex Form: [tex]\( y = 2(x + 1)^2 + 6 \)[/tex]
- Standard Form: [tex]\( y = 2x^2 + 4x + 8 \)[/tex]
These are the equations of the quadratic function that satisfy the given conditions.
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