Question

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

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.