3 years ago

Which of the following is the maximum value of the equation y=-2x^2-3x+4

1 answer:

4 0

Y=-2x^2-3x+4

dy/dx=-4x-3, d2y/dx2=-4

When the velocity, dy/dx=0, then y(x) is at an absolute maximum because the acceleration is a constant negative, -4.

dy/dx=0 only when -4x-3=0, -4x=3, x=-3/4

y(-3/4)=5.125

So the maximum value of y is 5 1/8

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