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Ilya [14]
3 years ago
15

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

Mathematics
1 answer:
mylen [45]3 years ago
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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Find the equation of the line through point (3,−3) and parallel to y=3x−4. Use a forward slash (i.e. "/") for fractions (e.g. 1/
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I'm fairly certain this is the answer.

Consider the point (3,-3). In terms of variables this is written (x,y).

Remember that the equation for slope is y = mx + b.

In that equation, "b" is your y-intercept, and "m" is your slope.

If 2 lines are parallel, they have the same slope but different y-intercepts. So you can use the slope from the equation y=3x-4 to find the parallel line. You can also plug in the values 3 and -3 to solve for b, our y-intercept.

If we set up the equation, we have -3 = 3(3) + b.

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