Graph the function y = f(x) = x².

If the graph of the given function were shifted six units to the left, the new graph would represent the function y = f(x + 6) = (x + 6)²y = f(x - 6) = (x - 6)²y = f(x) + 6 = x² + 6y = f(x) - 6 = x²- 6. If the graph of the given function were shifted upward by six units, the new graph would represent the function y = f(x + 6) = (x + 6)²y = f(x - 6) = (x - 6)²y = f(x) + 6 = x²+ 6y = f(x) - 6 = x²- 6. For the same y-values, the corresponding x-values of the function y = f(3x/2) = (3x/2)²y = f(2x/3) = (2x/3)²y = f(x + 1.5) = (x + 1.5)²y = f(x) + 1.5 = x² + 1.5 are 1.5 times of the x-values of the function y = f(x) = x2.

step-by-step explanation:

where is the question is there is a formula or is there not one

no oblique asymptote

step-by-step explanation:

to find the oblique asymptote of f(x)

here numerator has degree 1 and denomintor degree 2

degree of numerator < degree of denominator

hence there is no oblique asymptote

when the degree of the denominator > that of the numerator, we saw that the value in the denominator got so much bigger, so quickly,   giving us a horizontal asymptote of the x-axis. or we can write

f(x) = a polynomial of x + a proper fraction.

in this case only we have oblique asymptotes but not when numerator has less degree.

~dasgupta/mathbook/append17dec2k8.pdf