x+2y=12

find intercept
a circle equation x^2+y^2+4x-4y-1=0 and graph
2)find domain and range f(x)=x^2-4x-5
find vertex of f(x) =x^2+4x-21

(12,0) and (0,6)

center at (-2,2) radius 3
domain all reals, range all reals >= -9
(-2,-25)

To find the intercept of the equation x + 2y = 12, you can set either x or y to zero and solve for the other variable.

1. Setting x = 0:
Substituting x = 0 into the equation gives us: 0 + 2y = 12
Solving for y: 2y = 12
Dividing both sides by 2: y = 6
So the y-intercept is (0, 6).

2. Setting y = 0:
Substituting y = 0 into the equation gives us: x + 2(0) = 12
Simplifying: x = 12
So the x-intercept is (12, 0).

For the circle equation x^2 + y^2 + 4x - 4y - 1 = 0:

1. Rewrite the equation in standard form by grouping the x-terms and y-terms together:
x^2 + 4x + y^2 - 4y - 1 = 0

2. Complete the square for both x and y terms:
To complete the square for the x-terms, add (4/2)^2 = 4 to both sides:
x^2 + 4x + 4 + y^2 - 4y - 1 = 4
Simplifying: (x + 2)^2 + y^2 - 4y + 3 = 4

To complete the square for the y-terms, add (-4/2)^2 = 4 to both sides:
(x + 2)^2 + y^2 - 4y + 4 + 3 = 4 + 4
Simplifying: (x + 2)^2 + (y - 2)^2 = 5

Now, the equation of the circle is in standard form: (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center and r is the radius.
So, the center of the circle is (-2, 2) and the radius is sqrt(5).

To graph the circle, plot the center (-2, 2) and mark points on the graph that are the distance of sqrt(5) away from the center in all directions. Connect these points to form a circle.

For the function f(x) = x^2 - 4x - 5, to find the domain and range:

1. The domain represents all possible values of x for which the function is defined. In this case, since it is a quadratic function (x^2), the domain is all real numbers.

2. The range represents all possible values of y that the function can output. To find the range, we can analyze the concavity of the quadratic function. In this case, since the coefficient of x^2 is positive, the parabola opens upward.

To find the vertex of the function f(x) = x^2 - 4x - 5:

a. First, rewrite the quadratic function in the form f(x) = a(x - h)^2 + k.
f(x) = (x^2 - 4x) + (-5)
f(x) = (x^2 - 4x + 4 - 4) - 5
f(x) = (x - 2)^2 - 9

b. From the equation above, the vertex is at the point (2, -9).

Therefore, the domain of the function is all real numbers, and the range is y ≥ -9.