What is the #y#-intercept of #f(x) = 2(x-5)^2+12#?

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Answer 1 · 2018-04-13T08:53:16

#y#-intercept is #(0,62)#

For #y#-intercept one has to put #x=0# and for #x#-intercept one should put #y=0#

Here we have #y=2(x-5)^2+12#

henceputting #x=0# we get #y=2(0-5)^2+12#

or #y=2*25+12=62#

Hence, #y#-intercept is #(0,62)#

graph{2(x-5)^2+12 [-84.7, 75.3, -1, 79]}

Observe that we can never have #y=0# as minimum value of #y# is #12#, when #x=5#. Hence no #x#-intercept.