So, if we have the equation: F(x) = x^2 + 3 this is a function in terms of x, another way to look at this same problem is to write it as: y= x^2 +3. This function may be graphed if that is what you are looking for, the graph will be of a parabola and then the graph will be shifted from the origin up 3 from the origin.
f(x1) = (-5)2 + 3*(-5) + 5.1 = 25 - 15 + 5.1 = 15.1 f(x2) = (6)2 + 3*(6) + 5.1 = 36 + 18 + 5.1 = 59.1 f(x2)-f(x1) = 59.1 - 15.1 = 44 x2 - x1 = 6 - (-5) = 11 So average rate of change = 44/11 = 4
(x+3)(x+4) = x2+7x+12
At x = 3, the value of F(x) = 3x + 2 is the value 11, which graphs to the point (3, 11).
The value of x2+7x+10 depends on the value of X. If x=1, then x2+7x+10 = 18 If x=2, then x2+7x+10 = 28 If x=3, then x2+7x+10 = 40 and so forth
x2 + 3x - 18 = 0 (x - 3)(x + 6) = 0 x = 3 or -6
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If there is a remainder then it's not a factor otherwise yes
Are you trying to solve for x? Fx = x2 - 3 x2 - Fx - 3 = 0 x2 - Fx = 3 x2 - Fx + (F/2)2 = 3 + (F/2)2 (x - F/2)2 = 3 + (F/2)2 x - F/2 = ±[ 3 + (F/2)2 ]1/2 x = F/2 ± [ 3 + (F/2)2 ]1/2
x2 plus 3x plus 3 is equals to 0 can be solved by getting the roots.
f(x1) = (-5)2 + 3*(-5) + 5.1 = 25 - 15 + 5.1 = 15.1 f(x2) = (6)2 + 3*(6) + 5.1 = 36 + 18 + 5.1 = 59.1 f(x2)-f(x1) = 59.1 - 15.1 = 44 x2 - x1 = 6 - (-5) = 11 So average rate of change = 44/11 = 4
x2 + 10x + 21 = (x + 3)(x + 7)
In general, no.
y = 4x-3
x2 + 6x + 12 = 0 x2 + 6x + 9 = -3 (x + 3)2 = -3 x + 3 = ± √-3 x = -3 ± i√3
x2+12x+5 = 3 x2+12x+5-3 = 0 x2+12x+2 = 0
x2 + 4x + 3 = 0 (x + 1)(x + 3) = 0 x ∈ {-3, -1}
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