byndbtch
ax^2+bx+c<0 anyone have a clue as I dont!!
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Mertsj
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What are you supposed to do?
byndbtch
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solve the equation or inequality???
byndbtch
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there is a graph with it and I am so lost.....
byndbtch
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I had ax^2+bx+c=0 and got that one, I think but this????
Mertsj
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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.
byndbtch
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I know right....ok hang on. But I dont no how to post a graph??
Mertsj
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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.
byndbtch
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it is upward, the vertex is 6.5 and -4 and 4 does that help
byndbtch
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the vertex is -6
Mertsj
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|dw:1349043948873:dw|
Mertsj
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Does it look like that?
byndbtch
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kinda but the vertex ir right in the middle.....it points upward at -4 on the left and 4 on the right
Mertsj
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So the vertex is on the y axis?
byndbtch
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yes
Mertsj
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And it crosses the x axis at (4, 0) and (-4,0)?
byndbtch
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yes
byndbtch
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|dw:1349044244497:dw|
byndbtch
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more like that
Mertsj
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|dw:1349044229220:dw|
byndbtch
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exactly!!
Mertsj
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Now exactly what does the problem say to do?
byndbtch
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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
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There must be more because what you posted doesn't make sense.
byndbtch
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I know!! that is exactly as it is posted.
byndbtch
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my options for the second equation are like 7 choices...makes no sense to me??
Mertsj
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Does it say "The graph of y = x*2+bx+c is shown"?
byndbtch
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YES
Mertsj
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Ok. Now. I repeat my original request. Please post the problem EXACTLY as shown.
byndbtch
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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
byndbtch
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the vertex is not the -6 it is -30!!!!
byndbtch
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its -32
byndbtch
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I just seen that......does that help any???
Mertsj
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What are the x intercepts?
byndbtch
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they are the same
byndbtch
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-4 4
Mertsj
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Are you sure?
byndbtch
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yes
byndbtch
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|dw:1349045608333:dw|
Mertsj
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Then plug into the equation y = a(x-h)^2 +k and find the value of a
Mertsj
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Then solve the resulting inequality.
byndbtch
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that is where I am confused....plug what numbers? and plug them in as a????
Mertsj
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(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.
byndbtch
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ok I will try. thx
Mertsj
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yw