# Find the vertex, axis of symmetry, intercepts, domain, range, intervals where the function is increasing and decreasing. f(x)=x²-8x-4Please graph and show ALL work!!

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You need to remember that derivative of function tells you about the critical points of a function (if any) and it also describes the behaviour of the function, meaning if the function increases or decreases over specific intervals, hence you need to find derivative of function such that:

`f'(x) = 2x - 8`

You need to find the root of the equation f'(x) = 0, such that:

`2x - 8= 0 =gt x - 4 = 0 =gt x = 4`

Hence,the function has a critical value at x = 4. You need to find the critical point substituting 4 for x in the equation of function such that:

`f(4) = 4^2 - 8*4 - 4 =gt f(4) = 16 - 32 - 4 =gt f(4) = -20`

Hence, the function has a critical point at (4,-20).

You need to analyze if the function increases or decreases over specific intervals, hence you should analyze if the derivative function increases or decreases around the value x=4.

Notice that if x < 4, the values of derivative are negative, hence the function decreases over `(-oo,4).`

If x>4, the values of derivative are positive, hence the function increases over `(4,oo).`

Hence, the function reaches its minimum point at the vertex (4,-20).

The domain of the function consists of all the values of x for the function does exist, hence you notice that any real value satisfies the equation of the function and the domain is the set R.

Substituting each real value for x, yields also a real value, hence the range of the function is the set R.

The axis of symmetry of the quadratic function passes through the vertex (4,-20) and it is parallel to y axis, hence the line x=4 represents the axis of symmetry.

You need to remember that the graph of function intercepts x axis at y=0, hence you need to solve the equation f(x)=0 to find x intercepts.

`x^2 -8x - 4 = 0`

You need to apply quadratic formula to find `x_(1,2), ` hence:

`x_(1,2) = (8+-sqrt(64+16))/2 =gt x_1 = (8+4sqrt5)/2`

`x_1 = 4 + 2sqrt5 ; x_2 = 4 - 2sqrt5`

Hence, the graph intercepts x axis at `(4-2sqrt5,0)` and `(4+2sqrt5,0)` .

You should remember that the graph intercepts y axis at x = 0, hence you need to evaluate f(0)=-4.

Hence, the graph intercepts y axis at (0,-4).

The sketch of the graph below shows all the problem requests such that:

**Hence, the point (4,-20) denotes the vertex of function, the points `(4-2sqrt5,0)` and `(4+2sqrt5,0)` represent x intercepts, the point `(0,-4)` represents y intercept, the domain and the range of function are alike and represented by `R` , the axis of symmetry is `x=4` , and the function decreases over `(-oo,4)` and it increases over `(4,oo) ` .**