#1**0 **

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

Rewriting the equation

because I almost missed

the negative sign at x^{2} y = –x^{2} + 6x – 5

To find the zeros, then

set y equal to zero –x^{2} + 6x – 5 = 0

I recognize this as a

factorable quadratic

Factor out the –1 so

that it's easier to see (–1) * (x^{2} – 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**

_{.}

Guest Apr 27, 2023