precal question

Question

Evaluate the function at the indicated values. (If an answer is undefined, enter UNDEFINED.)

k(x) = −x² − 4x + 5;



k(0), k(4), k(−4), k(

4), k(a + 4), k(−x), k(x²)

Answer 1

k(x) = −x^2 − 4x + 5;  Find k(0), k(4), k(−4), k( 4), k(a + 4), k(−x), k(x2)

k(0) = -(0)^2 - 4(0) + 5 => 0 - 0 + 5 = 5

k(4) = -(4)^2 - 4(4) + 5 => -16 -16 + 6 => -32 + 5 = -28

k(-4) = -(-4)^2 -4(-4) + 5 => -16 + 16 + 5 = 5

k(a + 4) = -(a + 4)^2 - 4(a+4) + 5 => -(a^2 +8a +16) - 4a -16 + 5 => -a^2 -8a - 16 -4a -16 +6 => -a^2 -12a - 26

k(-x) = -(-x)^2 - 4(-x) + 5 => -x^2 + 4x + 5

k(x^2) = -(x^2)^2 - 4(x^2) + 5 => -x^4 -4x^2 + 5

Answer 2

k(x) is a continuous function, so it will be possible to evaluate it for all the given values.

For k (0), k(4), k(-4) substitute x = 0, 4, -4 into k(x) and evaluate

for k (√4) , conventionally we take the positive squareroot. So √4 = 2, so find k(2)

For k (a+4), k (-x) k (x²) for x in f(x) write a+4, -x, and x²

So f (x² ) = - (x²)² - 4 (x²) + 5 = -x⁴ - 4x² + 5

Mike