Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

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

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
What are you supposed to do?
solve the equation or inequality???
there is a graph with it and I am so lost.....

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question