f(x)= ⎨x^2-1 / x+1, for x not = -1

⎨8, for x=-1

I'm trying to graph and determine the values of x for which the function is continuous.

Does the graph looks like a u shape opening up or a line.

f(x) = (x^2 - 1)/(x+1)

= (x+1)(x-1)/(x+1)
= x-1 , x ≠ -1

so the graph will actually be a straight line
y = x-1 , with a hole at (-1, -2)

the graph will be continuous for all values of x, except x = -1

So the function is defined or not defined.

The question wasn't about a function being defined or not.

You wanted to know if it was continuous.

As I said, it is defined and continuous for all values of x , except x = -1

for all values of x, x ≠-1 , you have the function
f(x) = x-1, which is a straight line
the point (-1,-2) is a hole in the straight line and it is being replaced with (-1,8) , a single point lying above the line.
Strange function.

To determine the shape of the graph, we can analyze the equation of the function f(x).

For x ≠ -1, the function f(x) is given as (x^2 - 1) / (x + 1).
Note that the numerator (x^2 - 1) is a quadratic equation, and the denominator (x + 1) is linear.

When graphing a rational function, we usually look for horizontal and vertical asymptotes, as well as any holes in the graph.

Vertical Asymptotes:
Vertical asymptotes occur where the denominator equals zero and the numerator doesn't cancel out the zero. In this case, x + 1 = 0. Therefore, the vertical asymptote is x = -1.

Hole:
A hole appears in the graph if both the numerator and denominator share a common factor. In this case, (x - 1) is a common factor. Thus, there's a hole at x = 1, where the value of the function is undefined.

Horizontal Asymptote:
To find the horizontal asymptote, we examine the degrees of the numerator and denominator. The degree of the numerator is 2 (highest power of x), and the degree of the denominator is 1. Since the degree of the numerator is greater, there is no horizontal asymptote.

Based on this analysis, the graph of the function will have a vertical asymptote at x = -1 and a hole at x = 1. The shape of the graph will resemble a "U" shape opening up or down, depending on the values of x.

Now let's look at the function's values to determine where it is continuous. To find the values of x for which the function is continuous, we need to check if there are any breaks or discontinuities when x is equal to -1 or when x is not equal to -1.

At x = -1, the function is defined as f(-1) = 8, which means there is a continuous point at x = -1.

For x ≠ -1, the function can be evaluated as f(x) = (x^2 - 1) / (x + 1). Since the function is a rational function and we already addressed the vertical asymptote at x = -1, there are no other breaks or discontinuities.

Therefore, the function f(x) is continuous for all values of x except for the hole at x = 1.

To summarize, the graph of the function resembles a "U" shape opening up or down, and it is continuous for all values of x except for the hole at x = 1.