How do you find the vertex and intercepts for #f(x)=(x-1)^2-1#?

Question

Isaiah Anthony · Accepted Answer

Vertex (1, -1)

f(x) is written in vertex form. Vertex (1, -1) To find y intercept, make x = 0 --> y = 1 - 1 = 0 To find x-intercepts, make y = 0. #(x - 1)^2 - 1 = 0# #(x - 1)^2 = 1# x - 1 = +- 1 --> x = 1 + 1 = 2, and x = 1 - 1 = 0 graph{(x - 1)^2 - 1 [-5, 5, -2.5, 2.5]}

Ethan Adkins · Answer

undefined To find the vertex and intercepts of the function f(x) = (x - 1)^2 - 1, you can follow these steps:

1. **Vertex:** The vertex of a parabola in the form y = a(x - h)^2 + k is at the point (h, k). In this case, the function is f(x) = (x - 1)^2 - 1, which is in the form y = a(x - h)^2 + k with a = 1, h = 1, and k = -1. Therefore, the vertex is at (1, -1).

2. **Intercepts:**
   - **y-intercept:** To find the y-intercept, set x = 0 and solve for y: f(0) = (0 - 1)^2 - 1 = (-1)^2 - 1 = 1 - 1 = 0. So, the y-intercept is at (0, 0).
   - **x-intercepts:** To find the x-intercepts, set y = 0 and solve for x: 0 = (x - 1)^2 - 1. 
      - First, add 1 to both sides to get (x - 1)^2 = 1.
      - Then, take the square root of both sides (remembering to consider both the positive and negative square roots): x - 1 = ±1.
      - Finally, solve for x: x = 1 ± 1. This gives two solutions, x = 2 and x = 0. Therefore, the x-intercepts are at (2, 0) and (0, 0).

So, the vertex of the function is at (1, -1), the y-intercept is at (0, 0), and the x-intercepts are at (2, 0) and (0, 0).

Eva Abramson · Answer

undefined The vertex of the function f(x) = (x - 1)^2 - 1 is (1, -1). To find the x-intercepts, set f(x) equal to zero and solve for x. To find the y-intercept, evaluate f(x) at x = 0.