Question

F(x)=-1/2 x² - xfind the axis of symmetry, vertex, domain, and range.

Answer 1

Answer:

The axis of symmetry = -1

The vertex = (-1, 1/2)

The domain = all real numbers

The range = all real numbers less than or equal to 1/2

Explanations:

The given function is:

`f(x)\text{ = -}(1)/(2)x^2-x`

Note that for a function, f(x) = ax² + bx + c, the axis of symmetry is given by:

x = -b/2a

From the given function f(x):

a = -1/2, b = -1, c = 0

The axis of symmetry is therefore calculated as:

`\begin{gathered} x\text{ = }(-(-1))/(2((-1)/(2))) \\ x\text{ = -1} \end{gathered}`

Axis of symmetry: x = -1

The vertex of a quadratic equation is given as (h, k)

The axis of symmetry is the x-coordinate of the vertex, therefore, h = -1

To find k, let x = h and f(x) = k in the given equation

`\begin{gathered} k\text{ = -}(1)/(2)h^2-h \\ k\text{ =}-(1)/(2)(-1)^2-(-1) \\ k\text{ = -}(1)/(2)+1 \\ k\text{ = }(1)/(2) \\ \end{gathered}`

The vertex of the equation = (-1, 1/2)

Since f(x) is a quadratic function, the domain of the function f(x) is a set of all real numbers

To find the range:

Since a = -1/2 < 0, the graph is opening downwards.

Therefore, the range is given as y ≤ k

Since k = 1/2, the range is y ≤ 1/2