Which equation, in standard form, is equivalent to the equation below?

[tex]\[(y+7)^2 = -14(x-1)\][/tex]

A. [tex]\(x = \frac{1}{14} y^2 - y + 5\)[/tex]
B. [tex]\(x = \frac{1}{14} y^2 + y + 2\)[/tex]
C. [tex]\(x = -\frac{1}{14} y^2 - y + \frac{5}{2}\)[/tex]
D. [tex]\(x = -\frac{1}{14} y^2 - y - \frac{5}{2}\)[/tex]



Answer :

To determine which equation, in standard form, is equivalent to the given equation:

[tex]\[ (y + 7)^2 = -14(x - 1) \][/tex]

we follow these steps:

### Step 1: Expand the given equation

Start by expanding the left side of the equation [tex]\((y + 7)^2\)[/tex]:

[tex]\[ (y + 7)^2 = y^2 + 14y + 49 \][/tex]

So, the equation now reads:

[tex]\[ y^2 + 14y + 49 = -14(x - 1) \][/tex]

### Step 2: Distribute the right side

Next, distribute [tex]\(-14\)[/tex] on the right side of the equation:

[tex]\[ y^2 + 14y + 49 = -14x + 14 \][/tex]

### Step 3: Rearrange the equation to solve for [tex]\(x\)[/tex]

Add 14x to both sides to isolate [tex]\(x\)[/tex]:

[tex]\[ y^2 + 14y + 49 + 14x = 14 \][/tex]

Then, move constants to the right side:

[tex]\[ 14x = -y^2 - 14y + 14 - 49 \][/tex]

Combine like terms:

[tex]\[ 14x = -y^2 - 14y - 35 \][/tex]

Finally, divide each term by 14 to isolate [tex]\(x\)[/tex]:

[tex]\[ x = -\frac{1}{14} y^2 - y - \frac{35}{14} \][/tex]

Simplify the constant:

[tex]\[ x = -\frac{1}{14} y^2 - y - \frac{5}{2} \][/tex]

### Step 4: Check against the given standard form equations

We compare the derived equation with the given options:

1. [tex]\(x = \frac{1}{14} y^2 - y + 5\)[/tex]
2. [tex]\(x = \frac{1}{14} y^2 + y + 2\)[/tex]
3. [tex]\(x = -\frac{1}{14} y^2 - y + \frac{5}{2}\)[/tex]
4. [tex]\(x = -\frac{1}{14} y^2 - y - \frac{5}{2}\)[/tex]

Clearly, the derived equation matches the fourth given option:

[tex]\[ x = -\frac{1}{14} y^2 - y - \frac{5}{2} \][/tex]

### Conclusion

Thus, the equation in standard form that is equivalent to the given equation [tex]\((y + 7)^2 = -14(x - 1)\)[/tex] is:

[tex]\[ \boxed{-\frac{1}{14} y^2 - y - \frac{5}{2}} \][/tex]

Therefore, the answer is:

[tex]\[ x = -\frac{1}{14} y^2 - y - \frac{5}{2} \][/tex] which corresponds to option 4.

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