Question

I’m not sure if I did this right. How is this done step by step?

For each problem: Sketch the graph of: you must use algebra to find: _x and y intercepts, horizontal & vertical asymptotes, at least 3 additional points...and show the graph 1. f(x)- (x2-4)/(x2-3x +2) 2. h(x) (3x)/ (x+2x-3)

Answer 1

1) For x-intercept we will take f(x) = 0

(x^2 - 4)/(x^2 - 3x + 2) = 0

or, (x^2 - 4) = 0

or, (x + 2)(x - 2) = 0

x = -2, 2

So there is a x-intercept at (-2, 0) and (2, 0)

For y-intercept substitute x = 0

y = (0 - 4)/((0 - 0 + 2)

or, y = -4/2 = -2

So the intercept is at (0, -2)

For vertical asymptote,

x^2 - 3x + 2 = 0

or, x^2 - 2x - x + 2 = 0

or, x(x - 2) -1(x - 2) = 0

or, (x - 2)(x - 1) = 0

x = 1, 2

So the vertical asymptote is at x = 1 and x = 2

As the both numerator and denominator are 2nd degree polynomials, so for horizontal asymptote we must divide the coefficients of the highest terms.

So the horizontal asymptote is 1/1 = 1

2) For x-intercept take h(x) = 0

3x/(x^2 + 2x - 3) = 0

or, 3x = 0

or, x = 0

So there is a x-intercept at (0, 0)

For y-intercept substitute x = 0

h(x) = 0/(0 + 0 - 3)

h(x) = 0/(-3) = 0

So the y-intercept is at (0, 0).

For vertical asymptote

x^2 + 2x - 3 = 0

or, x^2 + 3x - x - 3 = 0

or, x(x + 3) -1(x + 3) = 0

or, (x + 3)(x - 1) = 0

x = -3, 1

So the vertical asymotote is at x = -3 and x = 1

The numerator cointains 1st degree polynomial and the denominator cointains 2nd degree polynomial.

Since the polynomial in numerator is a lower degree than the denominator, so the horizontal asymptote is located at y = 0.