Find the maximum or minimum value of function and the value of x when it occurs: -2x^2+5x+5.
let y = -x^2 + 5x + 5
dy/dx = -2x + 5
= 0 for a max of y
2x = 5
x = 5/2
when x = 5/2
y = -25/4 + 5(5/2) + 5
= 45/4
you should recognize y = -x^2 + 5x + 5 as a downward opening parabola ,
the max is 45/4 when x = 5/2
but the answer says that it's a max of 65/8 at x=5/4
Correct. Reiny missed the -2x^2. Even experts trip over the details sometimes.
so can you show me the correct way to do it?
wait nvm thanks for your help :D
To find the maximum or minimum value of a function, we need to determine the vertex of the quadratic equation. The vertex of a quadratic equation in the form of ax^2 + bx + c can be found using the formula:
x = -b / (2a)
Given the equation -2x^2 + 5x + 5, we can identify a = -2, b = 5, and c = 5. Plugging these values into the formula, we have:
x = -5 / (2(-2))
x = -5 / -4
x = 5/4 or 1.25
To find the maximum or minimum value of the function, we substitute the x-coordinate of the vertex back into the original equation. Plugging x = 1.25 into -2x^2 + 5x + 5, we get:
-2(1.25)^2 + 5(1.25) + 5
= -2(1.5625) + 6.25 + 5
= -3.125 + 6.25 + 5
= 8.125
Hence, the maximum or minimum value of the function is 8.125, and it occurs when x = 1.25.