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What are the vertex and range of y = x + 1 + 3?

(0, 4); - (0, 4); 3 ≤y <[infinity]
(-1,3); - O(-1, 3); 3 ≤y<[infinity]

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Final answer:

Assuming the intended function is y = x^2 + 1 (due to a typo), the vertex is (0, 1) and the range is [1, infinity), since this is a quadratic function opening upwards.

Step-by-step explanation:

The question is asking for the vertex and range of the function y = x + 1 + 3. This appears to be a typo as the function is not properly formatted. However, if we assume the function is actually y = x^2 + 1, then we can find the vertex and range of the quadratic function.

A quadratic function in the form y = ax^2 + bx + c has a vertex at the point (-b/2a, f(-b/2a)). For the function y = x^2 + 1, the vertex would be (0, 1). The range of a quadratic function opening upwards (which occurs when a is positive) is all the y-values greater than or equal to the y-coordinate of the vertex. Therefore, for y = x^2 + 1, the range would be [1, infinity).

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