inverse of y=x^2 -10x 27
\(y= x^{2} -10x+27 \\ y-2= x^{2} -10x+27-2 \\ y-2= x^{2} -10x+25 \\ y-2= (x-5)^{2} \\ x-5= \sqrt{y-2} \\ x=5+ \sqrt{y-2}\)Therefore, the inverse 0f y = x^2 - 10x + 27 is \(y = 5+ \sqrt{y-2}\)
