All that can be said is that F(x) is a quadratic function of x, with two real roots.
F(x) = 15x2 - 2.5 + 3 That's a quadratic or 2nd degree polynomial in x.
No, they are not equivalent.
To find f(-3), we substitute -3 into the function f(x) = x^2 + x: f(-3) = (-3)^2 + (-3) = 9 - 3 = 6
-1
All that can be said is that F(x) is a quadratic function of x, with two real roots.
F(x) = 15x2 - 2.5 + 3 That's a quadratic or 2nd degree polynomial in x.
The domain is the whole real line (or complex plane).
No, they are not equivalent.
To find f(-3), we substitute -3 into the function f(x) = x^2 + x: f(-3) = (-3)^2 + (-3) = 9 - 3 = 6
-1
If f(x) = 35/5 + 3 then its inverse is f(x) = 5/3*(x - 3).
f(x)=x3-3x2-5x+39=(x+3)(x2-6x+13) It has three roots. One of which is x=-3. Using the quadratic equation: x = (6 +/- √(-16))/2 x = (6 +/- 4i)/2 = (3 +/- 2i) so, x=-3, x=3+2i, or x=3-2i
f(x) = ax^2 + bx + c, where a != 0 (for obvious reason: it wouldn't be a quadratic function)
Using the remainder theorem:- f(x) = 4x3+6x2+3x+2 f(x) becomes f(-3/2) or f(-1.5) because the divisor is 2x+3 f(-1.5) = 4(-1.5)3+6(-1.5)2+3(-1.5)+2 = -5/2 or -2.5 So the remainder is -2.5
The answer to f x x-5 x plus 3 standard form is -6. This is a math problem.
If: 5f+3 = 28 then f = 5 because (5*5)+3 = 28