If 
is a real number, what is the maximum possible value of the expression 
?
+23254 The graph of this is a parabola that opens downward. There are at least three ways to find the maximum value.
1) You can graph it to find its vertex.
2) If you are in calculus, find the first derivative and set that equal to zero.
f'(t) = -2t + 8 ---> -2t + 8 = 0 ---> -2t = -8 ---> t = 4 --->
f(4) = -(4)² + 8(4) - 4 = -16 + 32 - 4 = 12
3) If you want to do this by algebra: find the vertex by completing the square:
f(t) = -t² + 8t - 4
f(t) + 4 = -t² + 8t Add 4 to both sides
f(t) + 4 = -(t² - 8t) Factor out the -1
f(t) + 4 -16 = -(t² - 8t + 16) Complete the square by dividing -8 by 2 and squaring the answer
f(t) - 12 = -(t - 4)²
f(t) = -(t - 4)² + 12 ---> vertex at (4, 12) ---> max value = 12
+23254 The graph of this is a parabola that opens downward. There are at least three ways to find the maximum value.
1) You can graph it to find its vertex.
2) If you are in calculus, find the first derivative and set that equal to zero.
f'(t) = -2t + 8 ---> -2t + 8 = 0 ---> -2t = -8 ---> t = 4 --->
f(4) = -(4)² + 8(4) - 4 = -16 + 32 - 4 = 12
3) If you want to do this by algebra: find the vertex by completing the square:
f(t) = -t² + 8t - 4
f(t) + 4 = -t² + 8t Add 4 to both sides
f(t) + 4 = -(t² - 8t) Factor out the -1
f(t) + 4 -16 = -(t² - 8t + 16) Complete the square by dividing -8 by 2 and squaring the answer
f(t) - 12 = -(t - 4)²
f(t) = -(t - 4)² + 12 ---> vertex at (4, 12) ---> max value = 12
