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what are the zeros to the quadric equation y=-x^2+6x-5

 Apr 27, 2023
 #1
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what are the zeros to the quadric equation y=-x^2+6x-5   

 

 

Rewriting the equation   

because I almost missed   

the negative sign at x2                     y  =  –x2 + 6x – 5   

 

To find the zeros, then

set y equal to zero                           –x2 + 6x – 5  =  0 

 

I recognize this as a   

factorable quadratic  

 

Factor out the –1 so  

that it's easier to see                        (–1) * (x2 – 6x + 5)  =  0 

 

You could've just multiplied 

both sides by –1, but you  

need to leave it there to show  

that the parabola opens down.           


Factor the quadratic                      (–1) * (x – 1) * (x – 5)  =  0    

 

So the zeros are at                          x = 1  and  x = 5    

.

 Apr 27, 2023
 #2
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Factor \(y=-x^2+6x-5\): \((5-x)(x-1)\) You should be able to figure out what values of x put that equation equal to zero.

 May 2, 2023

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