Question 1013975:  So how would I go about solving this quadratic equation:
 
y=-1x^2 - 10x
 
I know the axis of symmetry is -5 because x=-b/2a. When I try to find the y axis however, I get y=25. Is that right? if not, how would I solve this question?  Thanks for helping me! 
 Found 3 solutions by  josgarithmetic, macston, MathTherapy: Answer by josgarithmetic(39630)      (Show Source): 
You can  put this solution on YOUR website! That is ONE equation with TWO variables, so if that is all, then nothing to "solve".  Your "try to find the y axis" makes no sense.  The y-axis is the y-axis;  the   is not clear.  
 
 
Best you could do is to look for x and y intercepts, and with whatever skills you know up to now, find axis of symmetry, and the vertex.
 
 
Here is what you can do. 
  
  
  
meaning the x-axis intercepts are 0 and -10; which you can also call the "roots" of  .
 
 
The form of the equation   tells you that you have a vertical axis of symmetry; and it will occur in the exact middle between the x-axis intercepts, as YOU ALREADY FOUND,   to be this symmetry axis.
 
 
The vertex happens where  , so you find the y-coordinate using   
  
  
  
  
The vertex is  (-5,75), and this is a MINIMUM, based on, again, the form of the equation you have. 
 Answer by macston(5194)      (Show Source): 
You can  put this solution on YOUR website! It depends on what you want to know. 
. 
y=-x^2-10x 
y=(-1x)(x+10) 
. 
To find the roots, let y=0 
(-1x)(x+10)=0 
-1x=0 --OR-- x+10=0 
x=0 --OR-- x=-10  
Intercepts x axis at 0 and -10. 
. 
To find the x value at maximum(or minimum): 
Set first derivative=0 
. 
f(x)=-x^2-10x . (original equation) 
. 
f'(x)=-2x-10 . (first derivative) 
0=-2x-10 
10=-2x 
-5=x   
. 
The sign of the second derivative tells if  
it has a maximum or minimum (if negative,  
it has maximum, if positive it has minimum) 
f''(x)=-2 . (second derivative) 
is negative, so there is a maximum where x=-5. 
.  
Find y where x=-5: 
. 
f(x)=-x^2-10x 
f(-5)=-(-5)^2-10(-5)=-25=50=25 
The maximum is at the point (-5,25),  
crosses x-axis at -10 and 0 
. 
GRAPH: 
. 
  
  
 Answer by MathTherapy(10557)      (Show Source): 
You can  put this solution on YOUR website! So how would I go about solving this quadratic equation:
 
y=-1x^2 - 10x
 
I know the axis of symmetry is -5 because x=-b/2a. When I try to find the y axis however, I get y=25. Is that right? if not, how would I solve this question?  Thanks for helping me! 
No, that's incorrect! You just found the coordinates of the VERTEX, but you need the solutions/zeroes/roots, instead. 
To find the solutions/zeroes/roots, you simply set the equation equal to 0, as follows:  , and solve for x.  
You should get:    
  | 
 
  
 
 |   
 
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