the vertex of the graph of y=(x-1)^2-5 is
for y = (x-h)^2 + kthe vertex is h,kcomparingy=(x-1)^2-5 with y = (x-h)^2 + kwe have h = 1, k = -5so the vertex of y=(x-1)^2-5 is (1,-5)
Answer:
(1,-5)
Step-by-step explanation:
