The function f (x) = AX2 = B, f (1 = - 1), f (2) = 8. (1) find the value of AB, (2) f (5) F (x) = AX2 = B, that 2 is small 2, the computer can't type it out

The function f (x) = AX2 = B, f (1 = - 1), f (2) = 8. (1) find the value of AB, (2) f (5) F (x) = AX2 = B, that 2 is small 2, the computer can't type it out

f(x)=ax^2+b f(1)=-1 f(2)=8
Take (1. - 1) (2.8) into the solution of the equation
a+b=-1
4a+b=8
a=3,b=-4
F (x) = ax ^ 2 + B into a, B
F (x) = 3x ^ 2-4 when x = 5, f (5)
=71

If the value of F (x) tends to infinity, does f (x) have a limit?

No. this function is divergent

Find the limit of function f (x) = √ (xsquare + x) - x when x tends to negative infinity
The answer is positive infinity, I found that IQ began to decline

My brother came to teach you
First of all, if there is a root sign, you have to go to the root sign first
Multiply by a √ (x square + x) + X and divide by a √ (x square + x) + X
You get that the molecule is the square of x minus the root plus x minus the square of X
The denominator is √ (xsquare + x) + X. after sorting out, there is only one X left in the molecule
Then put an X square in the root sign of denominator, and the root sign becomes √ x square (1 + 1 / x), and the whole denominator is √ x square (1 + 1 / x) + X
Next, the square of the denominator is raised outside the root sign, and the denominator is - x √ (1 + 1 / x) + X (because x tends to be negative infinity)
Then the numerator and denominator are divided by an X, and the numerator is 1, and the denominator is - √ (1 + 1 / x) + 1
In the root sign, it's 1, because 1 / 2 of X, when x tends to negative infinity, it's 0, so the molecule is - 1 + 1, it's zero, and the molecule is 1, so the answer is positive infinity,
I can't help it if I still can't understand it, but I can't say if I can't give it to you~