3 years ago

The vertex of the graph of y=(x-1)^2-5 is

2 answers:

7 0

For y = (x-h)^2 + k
the vertex is h,k

comparing
y=(x-1)^2-5 with y = (x-h)^2 + k

we have h = 1, k = -5

so the vertex of y=(x-1)^2-5 is (1,-5)

lorasvet [3.4K]3 years ago

4 0

Answer:

(1,-5)

Step-by-step explanation:

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