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To find the vertex of the given parabola, we need to transform the equation [tex]\( x^2 - 16y - x - 12 = 0 \)[/tex] into the vertex form [tex]\( y = a(x - h)^2 + k \)[/tex], where [tex]\((h, k)\)[/tex] is the vertex. Here are the detailed steps to do this:
1. Rearrange the Equation:
Start by isolating the [tex]\( y \)[/tex]-term.
[tex]\[ x^2 - x - 16y - 12 = 0 \][/tex]
Move the [tex]\( y \)[/tex]-terms to the right side:
[tex]\[ x^2 - x = 16y + 12 \][/tex]
2. Complete the Square for the [tex]\( x \)[/tex]-terms:
To complete the square, we need to transform [tex]\( x^2 - x \)[/tex] into a perfect square trinomial.
Take half of the coefficient of [tex]\( x \)[/tex] (which is [tex]\(-1\)[/tex]), square it, and then add and subtract this value within the equation.
[tex]\[ \left(\frac{-1}{2}\right)^2 = \frac{1}{4} \][/tex]
Add and subtract [tex]\( \frac{1}{4} \)[/tex] on the left side:
[tex]\[ x^2 - x + \frac{1}{4} - \frac{1}{4} \][/tex]
This restructuring gives us:
[tex]\[ x^2 - x + \frac{1}{4} = 16y + 12 + \frac{1}{4} \][/tex]
3. Simplify the Equation:
The left-hand side can now be written as a perfect square:
[tex]\[ \left(x - \frac{1}{2}\right)^2 = 16y + 12 + \frac{1}{4} \][/tex]
Convert the constant term on the right side to a common denominator to combine:
[tex]\[ 16y + 12 + \frac{1}{4} = 16y + \frac{48}{4} + \frac{1}{4} = 16y + \frac{49}{4} \][/tex]
Rewriting that we get:
[tex]\[ \left(x - \frac{1}{2}\right)^2 = 16y + \frac{49}{4} \][/tex]
4. Solve for [tex]\( y \)[/tex]:
Move the 16 to isolate [tex]\( y \)[/tex]:
[tex]\[ 16y = \left(x - \frac{1}{2}\right)^2 - \frac{49}{4} \][/tex]
Divide through by 16:
[tex]\[ y = \frac{1}{16} \left( \left(x - \frac{1}{2}\right)^2 - \frac{49}{4} \right) \][/tex]
Simplify further:
[tex]\[ y = \frac{1}{16}(x - \frac{1}{2})^2 - \frac{49}{64} \][/tex]
Now, we can see the parabola's vertex form [tex]\( y = a(x - h)^2 + k \)[/tex] where [tex]\( h = \frac{1}{2} \)[/tex] and [tex]\( k = -\frac{49}{64} \)[/tex].
Therefore, the vertex of the parabola [tex]\( x^2 - 16y - x - 12 = 0 \)[/tex] is [tex]\(\left( \frac{1}{2}, -\frac{49}{64} \right) \)[/tex] or in decimal form:
[tex]\[ \left( 0.5, -0.765625 \right) \][/tex]
1. Rearrange the Equation:
Start by isolating the [tex]\( y \)[/tex]-term.
[tex]\[ x^2 - x - 16y - 12 = 0 \][/tex]
Move the [tex]\( y \)[/tex]-terms to the right side:
[tex]\[ x^2 - x = 16y + 12 \][/tex]
2. Complete the Square for the [tex]\( x \)[/tex]-terms:
To complete the square, we need to transform [tex]\( x^2 - x \)[/tex] into a perfect square trinomial.
Take half of the coefficient of [tex]\( x \)[/tex] (which is [tex]\(-1\)[/tex]), square it, and then add and subtract this value within the equation.
[tex]\[ \left(\frac{-1}{2}\right)^2 = \frac{1}{4} \][/tex]
Add and subtract [tex]\( \frac{1}{4} \)[/tex] on the left side:
[tex]\[ x^2 - x + \frac{1}{4} - \frac{1}{4} \][/tex]
This restructuring gives us:
[tex]\[ x^2 - x + \frac{1}{4} = 16y + 12 + \frac{1}{4} \][/tex]
3. Simplify the Equation:
The left-hand side can now be written as a perfect square:
[tex]\[ \left(x - \frac{1}{2}\right)^2 = 16y + 12 + \frac{1}{4} \][/tex]
Convert the constant term on the right side to a common denominator to combine:
[tex]\[ 16y + 12 + \frac{1}{4} = 16y + \frac{48}{4} + \frac{1}{4} = 16y + \frac{49}{4} \][/tex]
Rewriting that we get:
[tex]\[ \left(x - \frac{1}{2}\right)^2 = 16y + \frac{49}{4} \][/tex]
4. Solve for [tex]\( y \)[/tex]:
Move the 16 to isolate [tex]\( y \)[/tex]:
[tex]\[ 16y = \left(x - \frac{1}{2}\right)^2 - \frac{49}{4} \][/tex]
Divide through by 16:
[tex]\[ y = \frac{1}{16} \left( \left(x - \frac{1}{2}\right)^2 - \frac{49}{4} \right) \][/tex]
Simplify further:
[tex]\[ y = \frac{1}{16}(x - \frac{1}{2})^2 - \frac{49}{64} \][/tex]
Now, we can see the parabola's vertex form [tex]\( y = a(x - h)^2 + k \)[/tex] where [tex]\( h = \frac{1}{2} \)[/tex] and [tex]\( k = -\frac{49}{64} \)[/tex].
Therefore, the vertex of the parabola [tex]\( x^2 - 16y - x - 12 = 0 \)[/tex] is [tex]\(\left( \frac{1}{2}, -\frac{49}{64} \right) \)[/tex] or in decimal form:
[tex]\[ \left( 0.5, -0.765625 \right) \][/tex]
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