Question 666524
f(x) = {{{x^2 + 2x + 5}}}; evaluate f(a + 1)
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In f(a + 1), the a + 1 is taking the place of the x in f(x). As a result, simply plug the the a + 1 into the equation for f(x), exchanging each instance of x with a + 1: f(a + 1) = {{{(a + 1)^2 + 2(a + 1) + 5}}}
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Use the FOIL method to simplify {{{(a + 1)^2}}}: f(a + 1) = {{{a^2 + 2a + 1 + 2(a + 1) + 5}}}
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Distribute the 2 into the a + 1: f(a + 1) = {{{a^2 + 2a + 1 + 2a + 2 + 5}}}
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Simplify by adding together the 2a and 2a and the 1, 2, and 5: f(a + 1) = {{{a^2 + 4a + 8}}}
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Your final answer is f(a + 1) = {{{a^2 + 4a + 8}}}