Answer:
Step-by-step explanation:
![y=-x^2-12x-4\\\\y=-(x^2+12x)-4\\\\y=-(\underline{x^2+2\cdot x\cdot6+6^2}-6^2)-4\\\\y=-\left[(x+6)^2-36\right]-4\\\\y=-(x+6)^2+36-4\\\\\bold{\underline{y=-(x+6)^2+32}}](https://tex.z-dn.net/?f=y%3D-x%5E2-12x-4%5C%5C%5C%5Cy%3D-%28x%5E2%2B12x%29-4%5C%5C%5C%5Cy%3D-%28%5Cunderline%7Bx%5E2%2B2%5Ccdot%20x%5Ccdot6%2B6%5E2%7D-6%5E2%29-4%5C%5C%5C%5Cy%3D-%5Cleft%5B%28x%2B6%29%5E2-36%5Cright%5D-4%5C%5C%5C%5Cy%3D-%28x%2B6%29%5E2%2B36-4%5C%5C%5C%5C%5Cbold%7B%5Cunderline%7By%3D-%28x%2B6%29%5E2%2B32%7D%7D)
6c < 12
Divide by 6 on both sides:
c < 2
Answer : c< 2
1. Let the given line be the graph of function f.
2. Let g be the inverse function f.
3. If f(a)=b then g(b)=a. So if point (a, b) is on the original line, point (b, a) will be on its inverse.
4. Take a look at the choices we have.
In A, the first point is (-2, -11), which means that in the line we should have (-11, -2). Clearly we don't.
In C, point (-11, 2) tells us that we should have point (2, -11) in the line, but clearly 2 is matched to a value close to -1
In D, (2, -11) indicates that (-11, 2) is in the lone, but is not even close.
ANSWER: B
check: All (-2, 11), (-1, 8), (0, 5), (2, -1) are in the given line
The radius of the circle = 0-(-4) = 4 ( because it is tangent to the x axis and the center is at (-2,-4).)
a.) equation is (x + 2)^2 + (y + 4)^2 = 16
b) table of points
x y
-4 2
- 3 -0.13
-2 0
0 -0.54
and do the same for x = 1,2 and 3
