- Start with the standard form equation: \(y = 9x^2 + 9x - 1\).
- Factor out the leading coefficient from the first two terms to get \(y = 9(x^2 + x) - 1\).
- Complete the square by adding and subtracting \(\frac{1}{4}\) within the parentheses, resulting in \(y = 9(x + \frac{1}{2})^2 - \frac{9}{4} - 1\).
- Simplify the constant term to arrive at the final vertex form: \(\boxed{y=9\left(x+\frac{1}{2}\right)^2-\frac{13}{4}}\).
Explanation
- Problem Analysis
The given quadratic equation in standard form is \(y = 9x^2 + 9x - 1\). The standard form is generally written as \(y = ax^2 + bx + c\), where in this case, \(a = 9\), \(b = 9\), and \(c = -1\). The vertex form of a quadratic equation is given by \(y = a(x - h)^2 + k\), where \((h, k)\) is the vertex of the parabola. We will use the method of "completing the square" to transform the equation into this form.
- Completing the Square
To start, factor out the leading coefficient \(a = 9\) from the first two terms of the expression:
\[y = 9(x^2 + x) - 1\]
Next, identify the constant to add and subtract inside the parentheses to complete the square for the quadratic expression \(x^2 + x\). We take half of the coefficient of \(x\), which is \(1\), so half of it is \(\frac{1}{2}\). Squaring this gives \(\left(\frac{1}{2}\right)^2 = \frac{1}{4}\).
Add and subtract this value inside the parentheses:
\[y = 9\left(x^2 + x + \frac{1}{4} - \frac{1}{4}\right) - 1\]
- Simplification and Final Result
Distribute the \(9\) to the \(\frac{1}{4}\) term and move it outside the parentheses:
\[y = 9\left(x^2 + x + \frac{1}{4}\right) - 9\left(\frac{1}{4}\right) - 1\]
Now, rewrite the expression inside the parentheses as a squared binomial, \((x + \frac{1}{2})^2\):
\[y = 9\left(x + \frac{1}{2}\right)^2 - \frac{9}{4} - 1\]
Finally, simplify the constant terms by finding a common denominator:
\[-\frac{9}{4} - 1 = -\frac{9}{4} - \frac{4}{4} = -\frac{13}{4}\]
Thus, the vertex form of the quadratic equation is:
\[y = 9\left(x + \frac{1}{2}\right)^2 - \frac{13}{4}\]
Examples
Vertex form is essential in real-world applications such as physics and engineering, particularly when calculating the maximum or minimum point of a trajectory or a curve. For example, if you are modeling the path of a basketball shot or the suspension of a bridge, the vertex \((h, k)\) represents the highest or lowest point (the peak or the dip). By converting the equation of a trajectory from standard form to vertex form, you can instantly identify the peak height (\(k\)) and the horizontal distance (\(h\)) at which this height is achieved, allowing for quick design and analysis.