Question

For the function f(x) = −2(x + 3)2 − 1, identify the vertex, domain, and range.

The vertex is (3, −1), the domain is all real numbers, and the range is y ≥ −1.
The vertex is (3, −1), the domain is all real numbers, and the range is y ≤ −1.
The vertex is (−3, −1), the domain is all real numbers, and the range is y ≤ −1.
The vertex is (−3, −1), the domain is all real numbers, and the range is y ≥ −1.

Answer 1

"For the function f(x) = −2(x + 3)^2 − 1, identify the vertex, domain, and range. "

Note: You meant x^2, not x2. "^" indicates exponentiation.

This is a quadratic function of the form f(x) = ax^2 + bx + c. Its graph is a vertical parabola that opens down. We know this because a is negative 1 here, and the highest power of x is 2.

We can identify the vertex as being (-3, -1).

The domain of all quadratic functions is "the set of all real numbers."

You must find the maximum value of this function, which will represent the upper limit of the range (-infinity, y-value at vertex). Know how to do this? If not, please ask.

Answer 2

C. The vertex is
`(-3,-1)`, the domain is all real numbers, and the range is
`y\leq -1`.

Explanation:

We have been given a function
`f(x)=-2(x+3)^2-1`. We are asked to identify the vertex, domain and range of the given function.

We can see that our given parabola is in vertex form
`y=a(x-h)^2+k`, where
`(h,k)` represents vertex of parabola.

We can rewrite our given equation as:

`f(x)=-2(x-(-3))^2-1`

Therefore, the vertex of our given parabola would be
`(-3,-1)`.

We know that parabola is a quadratic function and the domain of a quadratic function is all real numbers.

We know that range of a quadratic function in form
`f(x)=a(x-h)^2+k` is:

`f(x)\leq k`, when
`a<0` and,

`f(x)\geq k`, when
`a>0`

Upon looking at our given function, we can see that
`a=-2`, which is less than 0, therefore, the range of our given function would be
`y\leq -1`.