+1207 If \(t\) is a real number, what is the maximum possible value of the expression \(-t^2 + 8t -4\)?
+129852 We have the form at^2 + bt + c where a = -1 and b = 8
This is a parabola that turns downward.....the y value of the vertex will be the max
The "t" value of the vertex = -b / [ 2a ] = - [ 8 ] / [2 (-1) ] = -8/ -2 = 4
So......the max value is -(4)^2 + 8(4) - 4 = -16 + 32 - 4 = -16 + 28 = 12
The vertex = ( t, y) = (4, 12)
Here's the graph showing this : https://www.desmos.com/calculator/6grvgwmlqj
+142 We can also complete the square. We can factor out a -1 from the given expression. We have -(t^2-8t+4). We can work inside the parenthesis. We can try to make the quadratic a perfect square. We have: t^2-8t+4=(t-4)^2-16+4=(t-4)^2-12. Remember the negative ;) We now have: -(t^2-4)^2+12. Since the negative of a positive number (a square) is always negative, the maximum it can be is 0. Therefore, the only possible maximum is 12. Hope this gives helpful insight :)
