Determine whether each function graphed or defined is one-to-one. y = 2(x+1)^2 - 6

Verified step by step guidance

1. Understand the definition of a one-to-one function: A function is one-to-one if each output value is paired with exactly one input value, meaning it passes the horizontal line test.
2. Consider the given function: \( y = 2(x+1)^2 - 6 \). This is a quadratic function, which typically graphs as a parabola.
3. Recall that a parabola opens upwards or downwards and is symmetric about its vertex. For a function to be one-to-one, it must pass the horizontal line test, meaning no horizontal line should intersect the graph more than once.
4. Identify the vertex of the parabola: The vertex form of a quadratic function is \( y = a(x-h)^2 + k \). Here, \( a = 2 \), \( h = -1 \), and \( k = -6 \), so the vertex is \((-1, -6)\).
5. Determine the direction of the parabola: Since \( a = 2 \) is positive, the parabola opens upwards. An upward-opening parabola is not one-to-one because horizontal lines will intersect it at two points, except at the vertex.

Key Concepts

One-to-One Function

A one-to-one function is a type of function where each output value is associated with exactly one input value. This means that no two different inputs produce the same output. To determine if a function is one-to-one, one can use the horizontal line test: if any horizontal line intersects the graph of the function more than once, the function is not one-to-one.

Quadratic Functions

Quadratic functions are polynomial functions of degree two, typically expressed in the form y = ax^2 + bx + c. They graph as parabolas, which can open upwards or downwards depending on the sign of the coefficient 'a'. Since parabolas are symmetric, they often fail the horizontal line test, indicating that they are not one-to-one unless restricted to a specific domain.

Vertex Form of a Quadratic

The vertex form of a quadratic function is given by y = a(x-h)^2 + k, where (h, k) is the vertex of the parabola. This form makes it easy to identify the vertex and the direction in which the parabola opens. In the given function, y = 2(x+1)^2 - 6, the vertex is at (-1, -6), and since it opens upwards, it confirms that the function is not one-to-one over its entire domain.