Let's go through the problem in your image step by step.

The given function:

$r(x) = \frac{8x - 8}{x + 2}$

1. Finding the x-intercept:

To find the x-intercept, set $r(x) = 0$ and solve for $x$.

$0 = \frac{8x - 8}{x + 2}$

The numerator must be 0: $8x - 8 = 0 \implies x = 1$

So, the x-intercept is $(1, 0)$, which is correct in your image.

2. Finding the y-intercept:

To find the y-intercept, set $x = 0$ and solve for $r(0)$.

$r(0) = \frac{8(0) - 8}{0 + 2} = \frac{-8}{2} = -4$

So, the y-intercept is $(0, -4)$. The correct coordinates are $(0, -4)$, but in the image, the y-intercept is entered as $(-4)$ which seems to be missing the correct $x = 0$.

The correct y-intercept is $(0, -4)$.

3. Vertical asymptote:

A vertical asymptote occurs when the denominator is 0. Set the denominator equal to 0: $x + 2 = 0 \implies x = -2$

So, there is a vertical asymptote at $x = -2$.

4. Horizontal asymptote:

To find the horizontal asymptote, we compare the degrees of the numerator and the denominator:

• The degree of the numerator is 1 (since $8x - 8$ is linear).
• The degree of the denominator is 1 (since $x + 2$ is linear).

When the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients. The leading coefficient in the numerator is 8, and in the denominator is 1, so the horizontal asymptote is: $y = \frac{8}{1} = 8$

Summary of answers:

• x-intercept: $(1, 0)$
• y-intercept: $(0, -4)$
• Vertical asymptote: $x = -2$
• Horizontal asymptote: $y = 8$

Would you like any further explanation or help with graphing this function?

Here are 5 related questions to extend your understanding:

1. How would you find oblique asymptotes for rational functions?
2. Can a rational function have more than one vertical asymptote?
3. What happens to the graph of the function near the vertical asymptote?
4. How would you find holes in the graph of a rational function?
5. How do the intercepts and asymptotes influence the shape of the graph?

Tip:

Always check both the numerator and the denominator for common factors before finding asymptotes—this will help you avoid missing holes in the graph!