The parabola will show the vertex in the format: y-k = (x-h)^2, where the vertex point
lies at (h, k).

let's first put it in "y =" standard format:

Since we cannot get a perfect square out of this, we complete the square: a=1, b=2, c=3
(b/2)^2 = (2/2)^2 = 1, so

So there's +2 leftover, since 3-1=2; so:

Now we'll subtract the 2 from both sides to show our vertex:

where our vertex (h, k) is at (-1, 2)
Y-axis symmetry=(r, theta)=(-r,theta)
-5-5cos(theta)=r
-r=5+5cos(theta)
no y-axis symmetry
x-axis symmetry=(r,theta)=(r,-theta)
cosine is an even function, so yes it is symmetric around x-axis
origin symmetry=(r,theta)=(-r,theta) or (r, theta+pi)
no, as there is no y-axis symmetry
Answer: The answer is B
Step-by-step explanation: 122 >7 (z+8) -5 (6+z). Then switch them, 7 (z+8) -5 (6+z) <122. Expanded: 7 (z+8) -5 (6+z) <122:v 2z+26
2z+26<122, subtract 26 from both sides: 2z+26-26<122-26
Simplify: 2z<96, divide both sides by 2: 2z over 2 < 96 over 2
and finally simplified: Z<48
HOPE THIS HELPED YOU OUT!
IF NOT SORRY!
Answer: A, 3 B, 2 C, 6
Step-by-step explanation:
So you just subsitute -3 for x
f(-3)=-(-3)+15
f(-3)=3+15
f(-3)=18
answer is C
18