Certainly! Let's complete the square to find the vertex of the given parabola.
The given equation is:
[tex]\[ x^2 - 16y - x - 12 = 0 \][/tex]
Step-by-Step Solution:
1. Isolate the 'y' term:
To work with [tex]\( y \)[/tex], first isolate the quadratic and linear terms involving [tex]\( x \)[/tex]:
[tex]\[ x^2 - x - 12 = 16y \][/tex]
2. Complete the square for the quadratic expression:
The quadratic expression here is [tex]\( x^2 - x \)[/tex]. To complete the square, we need to rewrite it in the form [tex]\((x - h)^2\)[/tex].
a. Identify the coefficients:
- The coefficient of [tex]\( x^2 \)[/tex] is 1.
- The coefficient of [tex]\( x \)[/tex] is -1.
b. Take half of the linear coefficient and square it:
- Half of [tex]\(-1\)[/tex] is [tex]\(\frac{-1}{2}\)[/tex]
- Squaring this value gives: [tex]\(\left(\frac{-1}{2}\right)^2 = \frac{1}{4}\)[/tex].
c. Add and subtract [tex]\(\frac{1}{4}\)[/tex] inside the equation:
[tex]\[ x^2 - x + \frac{1}{4} - \frac{1}{4} - 12 = 16y \][/tex]
d. Rewriting within the square form gives:
[tex]\[ (x - \frac{1}{2})^2 - \frac{1}{4} - 12 = 16y \][/tex]
3. Simplify the equation:
Combine the constants on the left-hand side:
[tex]\[ (x - \frac{1}{2})^2 - \frac{1}{4} - 12 = -\frac{1}{4} - 12 = - \frac{49}{4} \][/tex]
So the equation becomes:
[tex]\[ (x - \frac{1}{2})^2 - \frac{49}{4} = 16y \][/tex]
4. Isolate 'y':
Solve for [tex]\( y \)[/tex]:
[tex]\[ 16y = (x - \frac{1}{2})^2 - \frac{49}{4} \][/tex]
Then,
[tex]\[ y = \frac{1}{16}(x - \frac{1}{2})^2 - \frac{49}{4 \times 16} \][/tex]
[tex]\[ y = \frac{1}{16}(x - \frac{1}{2})^2 - \frac{49}{64} \][/tex]
5. Identify the vertex:
The general vertex form of a parabola is given by:
[tex]\[ y = a(x - h)^2 + k \][/tex]
By comparing, we get:
[tex]\[ h = \frac{1}{2}, \quad k = - \frac{49}{64} \][/tex]
Thus, the vertex of the parabola [tex]\(x^2 - 16y - x - 12 = 0\)[/tex] is:
[tex]\[ \left( \frac{1}{2}, - \frac{49}{64} \right) = \left( -0.5, 0.6875 \right) \][/tex]
Finally, the vertex coordinates are:
[tex]\[ (-0.5, 0.6875) \][/tex]
