Answer:
Step-by-step explanation:
We want to solve for x in
We group constants on the right to get:
We apply difference of two squares
Answer:
40
Step-by-step explanation:
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:
(C) f’(c) = 0 and f”(c) > 0
Step-by-step explanation:
A minimum occurs where the first derivative is 0 (the tangent line is horizontal), and the second derivative is positive (concave up). The simplest example of this is a positive parabola, like y = x², which has a relative minimum at its vertex.
Answer:
The correct quotient is 3.
I think Derek has mistaken to divide 
by 
.
Step-by-step explanation:
We have to check the quotient of 
.
Part A:
Now, .
= 
So, the correct quotient is 3.
Part B:
Derek says that the quotient is 
.
So, I think Derek has mistaken to divide by . (Answer)
