anonymous
  • anonymous
ax^2+bx+c<0 anyone have a clue as I dont!!
Mathematics
jamiebookeater
  • jamiebookeater
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Mertsj
  • Mertsj
What are you supposed to do?
anonymous
  • anonymous
solve the equation or inequality???
anonymous
  • anonymous
there is a graph with it and I am so lost.....

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anonymous
  • anonymous
I had ax^2+bx+c=0 and got that one, I think but this????
Mertsj
  • Mertsj
You need to post the entire problem including the graph. The only thing you posted in the general quadratic inequality. Its solution depends entirely upon the values of a, b, and c.
anonymous
  • anonymous
I know right....ok hang on. But I dont no how to post a graph??
Mertsj
  • Mertsj
Well then, you could tell what its shape is (I'm assuming it's a parabola), tell if it is concave upward or downward and give its vertex and x intercepts (if any). That would give us a pretty good idea of the problem.
anonymous
  • anonymous
it is upward, the vertex is 6.5 and -4 and 4 does that help
anonymous
  • anonymous
the vertex is -6
Mertsj
  • Mertsj
|dw:1349043948873:dw|
Mertsj
  • Mertsj
Does it look like that?
anonymous
  • anonymous
kinda but the vertex ir right in the middle.....it points upward at -4 on the left and 4 on the right
Mertsj
  • Mertsj
So the vertex is on the y axis?
anonymous
  • anonymous
yes
Mertsj
  • Mertsj
And it crosses the x axis at (4, 0) and (-4,0)?
anonymous
  • anonymous
yes
anonymous
  • anonymous
|dw:1349044244497:dw|
anonymous
  • anonymous
more like that
Mertsj
  • Mertsj
|dw:1349044229220:dw|
anonymous
  • anonymous
exactly!!
Mertsj
  • Mertsj
Now exactly what does the problem say to do?
anonymous
  • anonymous
The graph of y=ax^2+bx+c solve each equation . ax^2+bx+c=0 ax^2+bx+c<0 ax^2+bx+c>0
Mertsj
  • Mertsj
There must be more because what you posted doesn't make sense.
anonymous
  • anonymous
I know!! that is exactly as it is posted.
anonymous
  • anonymous
my options for the second equation are like 7 choices...makes no sense to me??
Mertsj
  • Mertsj
Does it say "The graph of y = x*2+bx+c is shown"?
anonymous
  • anonymous
YES
Mertsj
  • Mertsj
Ok. Now. I repeat my original request. Please post the problem EXACTLY as shown.
anonymous
  • anonymous
the graph of y=ax^2+bx+c is given below. Solve each equation or inequality. ax^2+bx+c=0 ax^2+bx+c<0 ax^2+bx+c>0
anonymous
  • anonymous
the vertex is not the -6 it is -30!!!!
anonymous
  • anonymous
its -32
anonymous
  • anonymous
I just seen that......does that help any???
Mertsj
  • Mertsj
What are the x intercepts?
anonymous
  • anonymous
they are the same
anonymous
  • anonymous
-4 4
Mertsj
  • Mertsj
Are you sure?
anonymous
  • anonymous
yes
anonymous
  • anonymous
|dw:1349045608333:dw|
Mertsj
  • Mertsj
Then plug into the equation y = a(x-h)^2 +k and find the value of a
Mertsj
  • Mertsj
Then solve the resulting inequality.
anonymous
  • anonymous
that is where I am confused....plug what numbers? and plug them in as a????
Mertsj
  • Mertsj
(h,k) is the vertex which in your case is (0,-32) You are given the x intercepts. Choose one of them, perhaps (4,0). Replace x with 4 and y with 0 and solve for a.
anonymous
  • anonymous
ok I will try. thx
Mertsj
  • Mertsj
yw

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