Respuesta :

nobillionaireNobley
[tex]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} [/tex]

Therefore, the inverse 0f y = x^2 - 10x + 27 is [tex]y = 5+ \sqrt{y-2} [/tex]

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