How do you graph #y =(x^2-3)/(x-1)#?

Question

Nathan Axel · Accepted Answer

See explanation

If you are dealing with questions at this level you know how to find the #x and y# intercepts
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Undefined at #x=1# as we have #y=(-2)/0# Thus we have vertical asymptotic behaviour
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#color(blue)("Investigating "x->1)#

If we have #y=((x^2+deltax)^2-3)/(x+deltax-1)# where #x=1#

Then we have #y=("negative")/("positive") <0#

x -> 0^+ #}lim_(x->0^+)(x^2-3)/(x-1)->k# where #k->-oo#

Conversely

If we have #y=((x^2-deltax)^2-3)/(x-deltax-1)# where #x=1#

Then we have #y=("negative")/("negative") >0#

x-> 0^- #}lim_(x->0^-)(x^2-3)/(x-1)->k# where #k->+oo#
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#color(blue)("Investigating "x->+oo)#

The temptation is to state the the const values become insignificant so we end up with #y=x^2/x->y=x#
THIS IS WRONG

If we actually divide the denominator into the numerator we get

#y=x+(x-3)/(x-1)#

Now when we take limits we end up with #y=oo+(oo)/(oo)#

Set #x=oo# gives #color(blue)(y=x+1)# as an oblique asymptote

Kennedy Adams · Answer

undefined To graph the equation y = (x^2-3)/(x-1), we can follow these steps:

1. Determine the domain of the function by finding the values of x for which the denominator (x-1) is equal to zero. In this case, x cannot be equal to 1.

2. Find the y-intercept by substituting x = 0 into the equation. y = (0^2-3)/(0-1) simplifies to y = -3.

3. Determine the x-intercepts by setting y = 0 and solving for x. In this case, (x^2-3)/(x-1) = 0 has no real solutions.

4. Analyze the behavior of the function as x approaches positive and negative infinity. As x approaches positive or negative infinity, the function approaches the line y = x.

5. Plot additional points by choosing various x-values and calculating the corresponding y-values using the equation.

6. Connect the plotted points smoothly to form the graph of the function.

The resulting graph will be a curve that approaches the line y = x as x approaches positive or negative infinity, with a vertical asymptote at x = 1.