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:
The line the foci will be graphed is?
y=0
The foci are located at about ?
(4.58, 0) and (-4.58, 0)
What type of lines will the directrices be?
Both vertical
The directrices are located at about?
x= 5.46 and x= -5.46
Step-by-step explanation:
<span>2r2 + 3s3 − r2 + 4t2 − r2
</span>
= 19
