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It is the straight line through the points (0, -1) and (1, 0).

Q: What is the graph of the function fx the quantity of x squared plus 3 x minus 4 all over x plus 4?

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None of the graphs that I can see!

2m^2 - 8 -First you should factor out a two. --> 2(m^2-4) -You now have something squared minus something else squared; You have m squared minus 2 squared. Whenever you have something squared minus something squared as you do in this case, there is a simple rule to remember: You can reduce that expression into the quantity of the square root of the first number or variable plus the square root of the second number or variable Times the quantity of the square root of the first number or variable minus the second number or variable squared. --> In the case of your expression: ----> 2(m+2)(m-2)<-----

âˆ« f'(x)/( q2f(x)2 - p2) dx = [1/(2pq)ln[(qf(x) - p)/(qf(x) + p)]

2x squared minus 4

No, unless "a" happens to be equal to 0, or to 1.

Related questions

To shift a funcion (or its graph) down "a" units, you subtract "a" from the function. For example, x squared gives you a certain graph; "x squared minus a" will give you the same graph, but shifted down "a" units. Similarly, you can shift a graph upwards "a" units, by adding "a" to the function.

x = 1

This expression factors as x -1 quantity squared.

None of the graphs that I can see!

In such cases, there is usually a discontinuity when the denominator is zero. In other words, solve for:x + 2 = 0

68.3

2m^2 - 8 -First you should factor out a two. --> 2(m^2-4) -You now have something squared minus something else squared; You have m squared minus 2 squared. Whenever you have something squared minus something squared as you do in this case, there is a simple rule to remember: You can reduce that expression into the quantity of the square root of the first number or variable plus the square root of the second number or variable Times the quantity of the square root of the first number or variable minus the second number or variable squared. --> In the case of your expression: ----> 2(m+2)(m-2)<-----

the x-axis... obviously! the x-axis... obviously!

x2 - x - 2 = 0(x - 2)(x + 1) = 0x = 2 and -1

âˆ« f'(x)/[f(x)âˆš(f(x)2 - a2)] dx = (1/a)arcses(f(x)/a) + C C is the constant of integration.

âˆ« f'(x)/âˆš(a2 - f(x)2) dx = arcsin(f(x)/a) + C C is the constant of integration.

âˆ« f'(x)/( q2f(x)2 - p2) dx = [1/(2pq)ln[(qf(x) - p)/(qf(x) + p)]