1. answer the following for the given quadratic function

f(x)=x^2+10x-11
a.what is the vertex (h,k) of f?
b. what is the axis of symmetry?
Please show work.

2. if f(x)=Inx, g(x)=e^5x, and h(x)=x^6
find the following
a. find (fog)(x)=
what is the domain of fog?
b.find (gof)(x)=
what is the domain of (gof)?
c.find (foh)(x)=
what is the domain of (foh)?
Please show work

1a. To find the vertex (h, k) of the quadratic function f(x) = x^2 + 10x - 11, we can use the formula h = -b/2a and substitute it into the equation to find k.

Here, a = 1 (coefficient of x^2) and b = 10 (coefficient of x).

h = -10 / (2 * 1) = -5

Now, substituting h back into the equation to find k:

k = (-5)^2 + 10(-5) - 11 = 25 - 50 - 11 = -36

Therefore, the vertex (h, k) of f is (-5, -36).

1b. The axis of symmetry for a quadratic function is the vertical line that passes through the vertex. In this case, the axis of symmetry is x = -5.

2a. (fog)(x) means we need to find f(g(x)). Given f(x) = ln(x) and g(x) = e^(5x), substituting g(x) into f(x) gives:

(fog)(x) = f(g(x)) = ln(e^(5x))

Since ln(x) and e^x are inverse functions, they cancel each other out. Therefore, (fog)(x) simplifies to:

(fog)(x) = 5x

The domain of fog is all real numbers since there are no restrictions on x.

2b. Similarly, (gof)(x) means we need to find g(f(x)). Given g(x) = e^(5x) and f(x) = ln(x), substituting f(x) into g(x) gives:

(gof)(x) = g(f(x)) = e^(5ln(x))

Using the property that e^(ln(x)) = x, we can simplify:

(gof)(x) = e^(ln(x^5)) = x^5

The domain of (gof) is also all positive real numbers since x^5 is defined for positive values of x.

2c. (foh)(x) means we need to find f(h(x)). Given f(x) = ln(x) and h(x) = x^6, substituting h(x) into f(x) gives:

(foh)(x) = f(h(x)) = ln((x^6))

The domain of (foh) is all positive real numbers, as ln(x) is defined for positive values of x.

1. a. To find the vertex (h, k) of the quadratic function f(x) = x^2 + 10x - 11, we can use the vertex formula.

The vertex formula states that the x-coordinate of the vertex is given by h = -b/2a, and the y-coordinate of the vertex is given by k = f(h).

In this case, a = 1, b = 10, and c = -11.

Using the formula, we have:
h = -b/2a = -10/(2*1) = -10/2 = -5.

To find k, we plug in h into the function f(x):
k = f(-5) = (-5)^2 + 10(-5) - 11 = 25 - 50 - 11 = -36.

So the vertex (h, k) of f(x) is (-5, -36).

b. The axis of symmetry is a vertical line through the vertex. For a quadratic function in the form f(x) = ax^2 + bx + c, the equation of the axis of symmetry is x = -b/2a.

In this case, the function is f(x) = x^2 + 10x - 11, so the axis of symmetry is x = -10/(2*1) = -5.

2. a. (fog)(x) means to substitute the function g(x) into function f(x).
(fog)(x) = f(g(x)) = f(e^5x).

The domain of fog is the set of all real numbers for which the composite function (fog)(x) is defined.

Since e^5x is defined for all real numbers, the domain of fog is also all real numbers.

b. (gof)(x) means to substitute the function f(x) into function g(x).
(gof)(x) = g(f(x)) = g(ln(x)).

The domain of (gof) is the set of all real numbers for which the composite function (gof)(x) is defined.

Since ln(x) is defined only for positive real numbers, the domain of (gof) is x > 0.

c. (foh)(x) means to substitute the function h(x) into function f(x).
(foh)(x) = f(h(x)) = f(x^6).

The domain of (foh) is the set of all real numbers for which the composite function (foh)(x) is defined.

Since x^6 is defined for all real numbers, the domain of (foh) is all real numbers.

1. To find the vertex (h, k) of the quadratic function f(x) = x^2 + 10x - 11, we can use the formula h = -b/2a and substitute the values of a and b from the equation.

a. The vertex (h, k) can be found using the formula:
h = -b/2a
In our case, a = 1 and b = 10, so we have:
h = -(10)/(2 * 1)
= -10/2
= -5
The x-coordinate of the vertex is -5.

To find the y-coordinate (k), substitute the x-coordinate into the equation:
f(-5) = (-5)^2 + 10(-5) - 11
= 25 - 50 - 11
= -36
The y-coordinate of the vertex is -36.

Therefore, the vertex of the quadratic function f(x) = x^2 + 10x - 11 is (-5, -36).

b. The axis of symmetry is a vertical line passing through the vertex of the quadratic function.
Using the x-coordinate of the vertex (-5), we can say the equation of the axis of symmetry is x = -5.

2. Given the functions f(x) = ln(x), g(x) = e^(5x), and h(x) = x^6, we will find the values of the composite functions (fog)(x), (gof)(x), and (foh)(x), as well as their respective domains.

a. (fog)(x) represents the composite function of f(g(x)). To find this, we need to substitute g(x) into f(x) and simplify:

(fog)(x) = f(g(x))
= ln(g(x))
= ln(e^(5x))
= 5x

So (fog)(x) simplifies to 5x.

The domain of (fog)(x) is the same as the domain of g(x), which is all real numbers.

b. (gof)(x) represents the composite function of g(f(x)). To find this, we need to substitute f(x) into g(x) and simplify:

(gof)(x) = g(f(x))
= g(ln(x))
= e^(5 * ln(x))
= x^5

So (gof)(x) simplifies to x^5.

The domain of (gof)(x) is the same as the domain of f(x), which is x > 0 (positive real numbers).

c. (foh)(x) represents the composite function of f(h(x)). To find this, we need to substitute h(x) into f(x) and simplify:

(foh)(x) = f(h(x))
= f(x^6)
= (x^6)^2 + 10(x^6) - 11
= x^12 + 10x^6 - 11

So (foh)(x) simplifies to x^12 + 10x^6 - 11.

The domain of (foh)(x) is the same as the domain of h(x), which is all real numbers.