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
.
