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Answer
The vertex of the function \( f(x) = (x-3)^2 - 1 \) is at \( (3, -1) \).
Solution
**Step 1:** Recognize that the function is given in vertex form, which is
\[
f(x) = (x-h)^2 + k.
\]
**Step 2:** Compare \( f(x)=(x-3)^2-1 \) with the vertex form. Here, it is clear that:
\[
h = 3 \quad \text{and} \quad k = -1.
\]
**Step 3:** Therefore, the vertex of the function is:
\[
(3, -1).
\]
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Simplify this solution The Deep Dive
The vertex of the quadratic function \( f(x)=(x-3)^{2}-1 \) is located at the point (3, -1). This means the parabola opens upwards, and the vertex is the lowest point on the graph. In essence, the vertex gives you the minimum value of the function! For any quadratic in vertex form, \( f(x) = a(x-h)^2 + k \), the vertex is simply the point (h, k). So, if you can identify those values, you're already halfway to sketching a graph of the function! How neat is that?
