How do you find #P(1/4)# if #P(x) = 4x^2-9x-1#?

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Answer 1 · 2017-03-13T20:31:34

See the entire solution process below:

To find #P(1/4)# we must substitute #color(red)(1/4)# for each occurrence of #color(red)(x)# in #P(x)#

#P(color(red)(x)) = 4color(red)(x)^2 - 9color(red)(x) - 1# becomes:

#P(color(red)(1/4)) = 4(color(red)(1/4))^2 - (9 xx color(red)(1/4)) - 1#

#P(color(red)(1/4)) = (4 xx color(red)(1/16)) - 9/4 - 1#

#P(color(red)(1/4)) = 4/16 - 9/4 - 1#

#P(color(red)(1/4)) = 1/4 - 9/4 - 1#

#P(color(red)(1/4)) = -8/4 - 1#

#P(color(red)(1/4)) = -2 - 1#

#P(color(red)(1/4)) = -3#