There are 2 options to solve that.
1. The first one is by derivatives.
f(x)=x^2+12x+36
f'(x)=2x+12
then you solve that for f'(x)=0
0=2x+12
x=(-6)
you have x so for (-6) solve the first equation, then you find y
y=(-6)^2+12*(-6)+36=(-72)
so the vertex is (-6, -72)
2. The second option is to solve that by equations:
for x we have:
x=(-b)/2a
for that task we have
b=12
a=1
x=(-12)/2=(-6)
you have x so put x into the main equation
y=(-6)^2+12*(-6)+36=(-72)
and we have the same solution: vertex is (-6, -72)
For next task, I will use the second option:
y=x^2-6x
x=(-b)/2a
for that task we have
b=(-6)
a=1
x=(6)/2=3
you have x so put x into the main equation
y=3^2+(-6)*3=(--9)
and we have the same solution: vertex is (3, -9)
Answer:
Quadrant 4
Step-by-step explanation:
The x value is positive so it is either in 1 or 4
The y value is negative so it is either 3 or 4
To meet both conditions, it must be 4
One of the most fundamental truths of (Euclidean) geometry is that the ratio of the circumference to the diameter of any circle is a constant, and that constant is called pi (denoted by π π). Let C be the (length of the) circumference of a circle, and let d be its diameter. Then, we must have: C d = π C d = π
Answer:
I'm not sure what the answer is, but try Photo math. it helps me so much with my equations, and it'll even tell you the steps to get the answer.
Divide 4/2 so you get 2= n over 4 then multiply 2 by 4 and then multiply n over 4 by 4 you’ll get 4*2=n, multiply 4 and 2.
n=8
