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byndbtch
 2 years ago
ax^2+bx+c<0 anyone have a clue as I dont!!
byndbtch
 2 years ago
ax^2+bx+c<0 anyone have a clue as I dont!!

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Mertsj
 2 years ago
Best ResponseYou've already chosen the best response.0What are you supposed to do?

byndbtch
 2 years ago
Best ResponseYou've already chosen the best response.0solve the equation or inequality???

byndbtch
 2 years ago
Best ResponseYou've already chosen the best response.0there is a graph with it and I am so lost.....

byndbtch
 2 years ago
Best ResponseYou've already chosen the best response.0I had ax^2+bx+c=0 and got that one, I think but this????

Mertsj
 2 years ago
Best ResponseYou've already chosen the best response.0You 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
 2 years ago
Best ResponseYou've already chosen the best response.0I know right....ok hang on. But I dont no how to post a graph??

Mertsj
 2 years ago
Best ResponseYou've already chosen the best response.0Well 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
 2 years ago
Best ResponseYou've already chosen the best response.0it is upward, the vertex is 6.5 and 4 and 4 does that help

byndbtch
 2 years ago
Best ResponseYou've already chosen the best response.0kinda but the vertex ir right in the middle.....it points upward at 4 on the left and 4 on the right

Mertsj
 2 years ago
Best ResponseYou've already chosen the best response.0So the vertex is on the y axis?

Mertsj
 2 years ago
Best ResponseYou've already chosen the best response.0And it crosses the x axis at (4, 0) and (4,0)?

Mertsj
 2 years ago
Best ResponseYou've already chosen the best response.0Now exactly what does the problem say to do?

byndbtch
 2 years ago
Best ResponseYou've already chosen the best response.0The 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
 2 years ago
Best ResponseYou've already chosen the best response.0There must be more because what you posted doesn't make sense.

byndbtch
 2 years ago
Best ResponseYou've already chosen the best response.0I know!! that is exactly as it is posted.

byndbtch
 2 years ago
Best ResponseYou've already chosen the best response.0my options for the second equation are like 7 choices...makes no sense to me??

Mertsj
 2 years ago
Best ResponseYou've already chosen the best response.0Does it say "The graph of y = x*2+bx+c is shown"?

Mertsj
 2 years ago
Best ResponseYou've already chosen the best response.0Ok. Now. I repeat my original request. Please post the problem EXACTLY as shown.

byndbtch
 2 years ago
Best ResponseYou've already chosen the best response.0the 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
 2 years ago
Best ResponseYou've already chosen the best response.0the vertex is not the 6 it is 30!!!!

byndbtch
 2 years ago
Best ResponseYou've already chosen the best response.0I just seen that......does that help any???

Mertsj
 2 years ago
Best ResponseYou've already chosen the best response.0What are the x intercepts?

Mertsj
 2 years ago
Best ResponseYou've already chosen the best response.0Then plug into the equation y = a(xh)^2 +k and find the value of a

Mertsj
 2 years ago
Best ResponseYou've already chosen the best response.0Then solve the resulting inequality.

byndbtch
 2 years ago
Best ResponseYou've already chosen the best response.0that is where I am confused....plug what numbers? and plug them in as a????

Mertsj
 2 years ago
Best ResponseYou've already chosen the best response.0(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.
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