SOLUTION: Identify the axis of symmetry. f(x) = x2 + 10x + 9

Algebra ->  Graphs -> SOLUTION: Identify the axis of symmetry. f(x) = x2 + 10x + 9      Log On


   



Question 953079: Identify the axis of symmetry.

f(x) = x2 + 10x + 9

Found 2 solutions by Alan3354, KMST:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Identify the axis of symmetry.

f(x) = x2 + 10x + 9
---------------
AOS is thru the vertex.
x = -b/2a

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29=x%5E2%2B10x%2B9
f%28x%29=x%5E2%2B10x%2B25-25%2B9
f%28x%29=%28x%5E2%2B10x%2B25%29-25%2B9
f%28x%29=%28x%5E2%2B10x%2B25%29-16
f%28x%29=%28x%2B5%29%5E2-16
You can see that
by adding 25 I "completed the square %28x%2B5%29%5E2=x%5E2%2B10x%2B25 .
Of course I add to also add -25 to compensate, so as not to change the equation.
(You may call that subtracting 25 , but I prefer to see it as adding a negative number).
You can also see that opposite values for %28x%2B5%29 give the same value to y .
In other words,
x%2B5=v and x%2B5=-v result in the same y=f%28x%29 for any value of v .
That means that x%2B5=0<--->x=-5 is the axis of symmetry.
Points that are a distance v to either side of that line,
either with x-%28-5%29=v<-->x%2B5=v ,
or with -5-x=v<-->5%2Bx=-v ,
have the same y=f%28x%29=v%5E2-16 .
So, for every v=x%2B5%3C%3E0 ,
there will be a point %22%28%22v%22%2C%22v%5E2-16%22%29%22 and
a point %22%28%22-v%22%2C%22v%5E2-16%22%29%22 ,
which is symmetrical, each being the reflection of the other point
to the other side of the line x=-5<-->x%2B5=0 .
There is only one point with x=-5<-->v=x%2B5=0 ,
and it has y=-16 . That is the point %22%28%22-5%22%2C%22-16%22%29%22 .
Since all the other points have y=v%5E2-16%3E0 ,
the point %22%28%22-5%22%2C%22-16%22%29%22 is the vertex and minimum of the graph of f%28x%29=x%5E2%2B10x%2B9 .