How do you identify the important parts of #y = (x – 1)^2 – 16# to graph it?

1 Answer
zoe
Sep 26, 2015

Minimum point #rArr# #(1, -16)#
#y-#intercept #rArr# #(0, -15)#

Explanation:

From the completed square, you already know that the minimum points are #(1, -16)#

To find the #y-#intercept, you need to expand the brackets to get it in the form of #ax^2 + bx+c# , where #c# is your #y-#intercept.

So,

#(x-1)(x-1) -16#
= #x^2-x-x+1-16#
= #x^2-2x+1-16#
= #x^2-2x+15#

So #15# is your #y-#intercept.

graph{(x-1)^2-16 [-9.495, 10.505, -17.76, -7.76]}

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