Question

For the function f(x) = –(x 1)^2 + 4, identify the vertex, domain, and range.

a. The vertex is (–1, 4), the domain is all real numbers, and the range is y ≥ 4.
b. The vertex is (–1, 4), the domain is all real numbers, and the range is y ≤ 4.
c. The vertex is (1, 4), the domain is all real numbers, and the range is y ≥ 4.
d. The vertex is (1, 4), the domain is all real numbers, and the range is y ≤ 4.

Answer 1

`f(x) = a(x-p)^(2) +q \\ \\ W=(p,q)`

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`f(x) = -(x+1)^(2) +4 \\ \\ W=(-1,4)`

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`D: x \in R`

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`a \ \textless \ 0 \\ \\ R: y \leq 4`

Answer 2

the vertex is at -b/(2A). if you use the foil method it will come out to -(x^2+2x+1)+4. then distribute the negative and simplify the equation (add the four) the equation comes out to be -x^2-2x+3. The vertex is at x=2/(2*-1), which is -1. plug -1 into the original equation to find y, (y=-(-1)^2-2*(-1)+3) which is y=4. so the vertex is at (-1,4). then since the parabola opens down (negative a value) and has a maximum at 4, y is always less than or equal to 4. x is all real numbers as it goes on forever. so b is your answer.