What is the function $f(x)=x^2-2x-6$ written in vertex form?
To determine the vertex form of the quadratic function \(f(x) = x^2 - 2x - 6\), we need to rewrite it in the form \(f(x) = a(x - h)^2 + k\), where \((h, k)\) represents the vertex of the parabola. The standard form provided is \(f(x) = ax^2 + bx + c\) with \(a = 1\), \(b = -2\), and \(c = -6\). To transform this into vertex form, we use the technique of completing the square.
First, group the \(x\)-terms of the quadratic function:
\[f(x) = (x^2 - 2x) - 6.\]
To complete the square for the quadratic trinomial \(x^2 - 2x\), we take half of the coefficient of \(x\), which is \(-2/2 = -1\), and square it to get \((-1)^2 = 1\). We then add and subtract this value inside the function expression to maintain equality:
\[f(x) = (x^2 - 2x + 1) - 1 - 6.\]
Now, rewrite the perfect square trinomial \(x^2 - 2x + 1\) as \((x - 1)^2\) and simplify the remaining constants:
\[f(x) = (x - 1)^2 - 1 - 6\]
\[f(x) = (x - 1)^2 - 7.\]
Comparing this result to the given options, we find that the vertex form of the function is \(f(x) = (x - 1)^2 - 7\). The vertex of this parabola is at \((1, -7)\).
Quadratic functions often describe the trajectories of projectiles in real-world scenarios. For example, if you throw a ball, its height over time follows a parabolic path. Converting a quadratic equation to vertex form, like we did with \(f(x) = (x - 1)^2 - 7\), helps you quickly identify the maximum or minimum height reached (the vertex) and the time at which it occurs, making it a very useful tool for physics and engineering.