SOLUTION: I need to know what a parabola graph looks like when the parabola will cross the x-axis at 0 and 12. I worked out the problem I just need to know what the graph will look like. I d

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: I need to know what a parabola graph looks like when the parabola will cross the x-axis at 0 and 12. I worked out the problem I just need to know what the graph will look like. I d      Log On


   

Question 763418: I need to know what a parabola graph looks like when the parabola will cross the x-axis at 0 and 12. I worked out the problem I just need to know what the graph will look like. I don't understand that part of it.
My problem was: -25x^2 + 300x = 0
25x^2 - 300x = 0
25x (x - 12) = 0
25x = 0 or x -12 = 0
I just need to know how to graph it and see what it looks like. I believe the parabola will open downward because it has a negative value.
Found 2 solutions by ramkikk66, josmiceli:
Answer by ramkikk66(644) About Me  (Show Source):
You can put this solution on YOUR website!
Your solution is right. The quadratic has two roots x = 0 and x = 12.
See the graph below (I have divided the eqn by 25 to make -x^2 + 12*x = 0, which has the same 2 roots, to make it fit.) You are right that the graph opens downwards. It has a max value for x = 6.

:)
graph%28+600%2C+600%2C+-20%2C+20%2C+-50%2C+50%2C+-x%5E2+%2B+12%2Ax+%2B+0%29

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Here's the graph:
+graph%28+400%2C+400%2C+-4%2C+16%2C+-100%2C+1000%2C+-25x%5E2+%2B+300x+%29+
--------------
One thing to notice right away is it must go through
( 0,0 ) since making +x+=+0+ will force +y+=+0+
--------------
The x-crossings ( the roots ) are at ( 0,0 ) and ( 12,0 )
--------------
The vertex is at +-b%2F%282a%29+ when the equation has the form
+y+=+ax%5E2+%2B+bx+%2B+c+
For your equation:
+a+=+-25+
+b+=+300+
+c+=+0+
+-b%2F%282a%29+=+-300+%2F+%28+2%2A%28-25%29%29+
+-b%2F%282a%29+=+-300+%2F+%28-50%29+
+-b%2F%282a%29+=+6+
---------------
So, the vertex is at ( 6,y ) Now find +y+
+y+=+-25%2A6%5E2+%2B+300%2A6+
+y+=+-900+%2B+1800+
+y+=+900+
The vertex is at ( 6,900 )
The vertex is above the x-crossings, so it
must open down.