Sure, let's go through the problem step by step.
### Given Equations:
- Equation A: [tex]\( y = 3x^2 - 6x + 21 \)[/tex]
- Equation B: [tex]\( y = 3x^2 - 6x + 18 \)[/tex]
- Equation C: [tex]\( y = 3(x-1)^2 + 18 \)[/tex]
- Equation D: [tex]\( y = 3(x-1)^2 + 21 \)[/tex]
### Step 1: Simplify and Compare Equations
Let us identify which equations are equivalent by comparing their expanded form.
#### Simplifying Equation C:
[tex]\[ y = 3(x-1)^2 + 18 \][/tex]
Expanding:
[tex]\[ y = 3(x^2 - 2x + 1) + 18 \][/tex]
[tex]\[ y = 3x^2 - 6x + 3 + 18 \][/tex]
[tex]\[ y = 3x^2 - 6x + 21 \][/tex]
This matches with Equation A:
[tex]\[ \text{Equation C} \equiv \text{Equation A} \][/tex]
#### Simplifying Equation D:
[tex]\[ y = 3(x-1)^2 + 21 \][/tex]
Expanding:
[tex]\[ y = 3(x^2 - 2x + 1) + 21 \][/tex]
[tex]\[ y = 3x^2 - 6x + 3 + 21 \][/tex]
[tex]\[ y = 3x^2 - 6x + 24 \][/tex]
This does not match with any other equation.
### Step 2: Identify the Most Useful Form for Extreme Values
The vertex form (which makes it easy to find the vertex—thus the extreme value) of a quadratic equation is generally given by:
[tex]\[ y = a(x-h)^2 + k \][/tex]
Where [tex]\((h, k)\)[/tex] is the vertex of the parabola.
From the given equations:
- Equation C: [tex]\( y = 3(x-1)^2 + 18 \)[/tex] is in vertex form.
Checking for other equations in vertex form:
- Equation D: Although it is in vertex form, we already determined it is not equivalent to any of the other equations as it expands to [tex]\( 3x^2 - 6x + 24 \)[/tex].
### Summary:
1. Equivalent Equations:
- Equations A and C are equivalent.
2. Form Most Useful for Identifying the Extreme Value:
- Among the equations, C and D are in vertex form. However, since Equation D is not equivalent to any other, Equation C is the one that's useful for identifying the extreme value.
### Conclusion:
Equations [tex]\( \boxed{\text{A and C}} \)[/tex] are equivalent, and of those, equation [tex]\( \boxed{\text{C}} \)[/tex] is in the form most useful for identifying the extreme value of the function it defines.
