- Begin with the original vertex form equation: \(y = a(x - h)^2 + k\).
- Isolate \(a\) by subtracting \(k\) from both sides to get \(y - k = a(x - h)^2\).
- Divide both sides by \((x - h)^2\) to isolate \(a\), resulting in \(a = \frac{y - k}{(x - h)^2}\).
- Identify this as the correct equivalent equation from the given options.
- The final answer is \(\boxed{a = \frac{y - k}{(x - h)^2}}\).
Explanation
- Analysis of the Given Equation
The given equation is \(y = a(x - h)^2 + k\), which is in the vertex form of a quadratic function. Our objective is to determine which of the provided equations is algebraically equivalent to this formula. Algebraic equivalence means that we can rearrange the original equation to match one of the given choices using standard algebraic operations.
- Evaluating the Options
We examine the relationship \(y = a(x - h)^2 + k\) and isolate each variable to check the options:
- First, isolate \(a\):
Subtract \(k\) from both sides:
\[y - k = a(x - h)^2\]
Divide both sides by \((x - h)^2\) (assuming \(x
eq h\)):
\[a = \frac{y - k}{(x - h)^2}\]
This matches the fourth option.
- Let's verify why the other options do not match:
- For the first option, isolating \(x\):
\[(x - h)^2 = \frac{y - k}{a}\]
\[x - h = \pm \sqrt{\frac{y - k}{a}}\]
\[x = h \pm \sqrt{\frac{y - k}{a}}\]
This is not equal to \(x = \pm \sqrt{\frac{y - k}{a}} - h\).
- For the second option, \(k = y + (x - h)^2\) is clearly incorrect because the original equation implies \(k = y - a(x - h)^2\).
- For the third option, isolating \(h\):
\[x - h = \pm \sqrt{\frac{y - k}{a}}\]
\[h = x \mp \sqrt{\frac{y - k}{a}}\]
This is not equal to \(h = x - \left(\frac{y - k}{a}\right)^2\).
- Conclusion and Final Answer
Through algebraic manipulation, we have demonstrated that the only expression equivalent to the original equation \(y = a(x - h)^2 + k\) is \(a = \frac{y - k}{(x - h)^2}\). Therefore, the correct choice is the fourth option. The final answer is \(\boxed{a = \frac{y - k}{(x - h)^2}}\).
Examples
In physics, the vertex form of a quadratic equation is often used to describe the trajectory of a projectile. If you know the vertex \((h, k)\) and a point \((x, y)\) on the path, you can use the equivalent form \(a = \frac{y - k}{(x - h)^2}\) to solve for the acceleration constant \(a\). This calculation helps in predicting the peak height and the horizontal distance covered by an object launched at different angles.