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

- anonymous

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

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- Mertsj

What are you supposed to do?

- anonymous

solve the equation or inequality???

- anonymous

there is a graph with it and I am so lost.....

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- anonymous

I had ax^2+bx+c=0 and got that one, I think but this????

- 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

I know right....ok hang on. But I dont no how to post a graph??

- 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

it is upward, the vertex is 6.5 and -4 and 4 does that help

- anonymous

the vertex is -6

- Mertsj

|dw:1349043948873:dw|

- Mertsj

Does it look like that?

- anonymous

kinda but the vertex ir right in the middle.....it points upward at -4 on the left and 4 on the right

- Mertsj

So the vertex is on the y axis?

- anonymous

yes

- Mertsj

And it crosses the x axis at (4, 0) and (-4,0)?

- anonymous

yes

- anonymous

|dw:1349044244497:dw|

- anonymous

more like that

- Mertsj

|dw:1349044229220:dw|

- anonymous

exactly!!

- Mertsj

Now exactly what does the problem say to do?

- 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

There must be more because what you posted doesn't make sense.

- anonymous

I know!! that is exactly as it is posted.

- anonymous

my options for the second equation are like 7 choices...makes no sense to me??

- Mertsj

Does it say "The graph of y = x*2+bx+c is shown"?

- anonymous

YES

- Mertsj

Ok. Now. I repeat my original request. Please post the problem EXACTLY as shown.

- 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

the vertex is not the -6 it is -30!!!!

- anonymous

its -32

- anonymous

I just seen that......does that help any???

- Mertsj

What are the x intercepts?

- anonymous

they are the same

- anonymous

-4 4

- Mertsj

Are you sure?

- anonymous

yes

- anonymous

|dw:1349045608333:dw|

- Mertsj

Then plug into the equation y = a(x-h)^2 +k and find the value of a

- Mertsj

Then solve the resulting inequality.

- anonymous

that is where I am confused....plug what numbers? and plug them in as a????

- 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

ok I will try. thx

- Mertsj

yw

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