Q: Do you find the maximum value on the x or y-axis?

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Assuming the standard x and y axes, the range is the maximum value of y minus minimum value of y; and the domain is the maximum value of x minus minimum value of x.

In Calculus, to find the maximum and minimum value, you first take the derivative of the function then find the zeroes or the roots of it. Once you have the roots, you can just simply plug in the x value to the original function where y is the maximum or minimum value. To know if its a maximum or minimum value, simply do your number line to check. the x and y are now your max/min points/ coordinates.

Suppose the revenue equation is of the form R = ax2 + bx + c where a, b and c are constants and x is the variable. To have a maximum, either a must be negative or x must lie within fixed limits. If a is negative then the maximum revenue is attained when x = -b/(2a). That is, find the value of R when x = -b/(2a). If a is positive, then find the value of R when x is at each end point of its domain. One of them will be larger and that is the maximum value of the revenue.

Both the function "cos x" and the function "sin x" have a maximum value of 1, and a minimum value of -1.

up and down. the x goes left and right

If x is the unknown or variable in an equation it can have many possible maximum or minimum values

1

Sin(x) has a maximum value of +1 and a minimum value of -1.

If: 4x-6 = 54 then the value of x works out as 15

The only variable on the right hand side is sin(x). The maximum value of sin(x) is 1. So, the max value of 3sin(x) is 3*1 = 3 and so, the max value of 3sin(x) + 2 is 3+2 = 5.

18

The formula to find the value of X would be Y-2X. This would equal to y-9 times 2 X.

You look for the value of 0 in the y column, and find out what x has to be for y=0. This value of x is you x-axis intercept. (Reverse "x" and "y" in the above description to find the y-intercept, if there is one).

im trying to find the value of my x-men cards

8

.75/x=.33

42

The answer will depend on what x is.

You cannot. The function f(x) = x2 + 1 has no real zeros. But it does have a minimum.

Lenght x r x pie x 2

180-x...... hahaahahaaa

There is no numerical value for 'x' that can make this a true statement.

In theory you can go down the differentiation route but because it is a quadratic, there is a simpler solution. The general form of a quadratic equation is y = ax2 + bx + c If a > 0 then the quadratic has a minimum If a < 0 then the quadratic has a maximum [and if a = 0 it is not a quadratic!] The maximum or minimum is attained when x = -b/2a and you evaluate y = ax2 + bx + c at this value of x to find the maximum or minimum value of the quadratic.

x=100

Lenght x r x pie x 2

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