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

Answer 1

Nathan Axel

#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.

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Answer 2

Julia Adams

To find the y-intercept of the function ( f(x) = 2(x-5)^2 + 12 ), we set ( x ) to zero and solve for ( f(x) ):

[ f(x) = 2(0-5)^2 + 12 ] [ f(x) = 2(-5)^2 + 12 ] [ f(x) = 2(25) + 12 ] [ f(x) = 50 + 12 ] [ f(x) = 62 ]

Therefore, the y-intercept of the function ( f(x) = 2(x-5)^2 + 12 ) is ( y = 62 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.