y = 2x2 + 4x - 3
This equation describes a parabola. Because it's first term is positive, we know that it goes infinitely upward, and has a minimum that occurs at it's vertex. You can find it's vertex by taking it's derivative and solving for zero:
y' = 4x + 4
0 = 4x + 4
0 = x + 1
x = -1
y = (-1)2 + 4(-1) - 3
y = 1 - 4 - 3
y = -6
So the vertex is at (-1, -6), which means that y ≥ -6
4x3+10x2+2x+5= (2x+5)(2x2+1)
14
4x3=12 250-12=238 238/50=4.76
4x3+10x2-6x 2x(2x2+5x-3) 2x(2x-1)(x+3)
12 divided by 3 equals 4. Then, 4 times 3 equals 12. Finally, 4 plus 12 equals 16.
4x3+10x2+2x+5= (2x+5)(2x2+1)
2x2x3=12 so therefore 12=2x2x3 2x2=4x3=12
4x3 + 2x2 - 6 has three terms.
14
4x3=12 250-12=238 238/50=4.76
You can work this out with long division, by checking to see if (x2 - 1) is a factor of (2x4 + 4x3 - x2 + 4x - 3). It is. Unfortunately, the WikiAnswers system is somewhat limited in depicting things such as long division, so we won't be able to represent it here. In short though, (2x4 + 4x3 - x2 + 4x - 3) / (x2 + 1) is equal to 2x2 + 4x - 3. which means that: (x2 + 1) / (2x4 + 4x3 - x2 + 4x - 3) = (x2 + 1) / (x2 + 1)(2x2 + 4x - 3) = 1 / (2x2 + 4x - 3)
There is no maximum since a quartic with a positive leading coefficient increases without limit. The minimum is approximately -4.4940
The "degree" is the highest power - in this case, the 3 in 4x3 (4 times to the third power).
(4x3)+2 = 14
4x3+10x2-6x 2x(2x2+5x-3) 2x(2x-1)(x+3)
4x3+12x2+3x+9
∫(4x3 - 2x2 + x - 1) dx You can integrate this by taking the antiderivative of each term. Each of these terms is in the format axn, the antiderivative of which is axn-1/n: = ∫(4x3)dx - ∫(2x2)dx + ∫(x)dx - ∫(1)dx = x4 - 2x3/3 + x2/2 - x + C