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1. $Y$ intercept at $-5$
2. No $x$ intercepts
3. Discontinuous points at $(-1,-5)$ and $(3, -5)$

This was on an assignment, please help! Edit: the graph is NOT linear

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3 Answers

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Hint: Discontinuous points come from common factors in the numerator and denominator, so you should be able to find two factors of the denominator from item 3. Since you have no $x$ intercepts, the function should not go through zero, and it sounds like you don't want vertical asymptotes. It looks like a nice graph that satisfies your needs would be the line $y=-5$ less the two discontinuous points.

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$$y=\left(\frac{x+1}{\left| x+1\right|}\right)\left(\frac{x-3}{\left|x-3\right|}\right)-4$$

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$$y=\sin(2\pi x)-7+\left(\frac{x+1}{x+1}\right)+\left(\frac{x-3}{x-3}\right)$$

Explanation: $\sin$ term to make it non-linear and no x intercept

$-7$ to account for y intercept

Fractional terms to account for discontinuities (You can only simplify those to $1$ if $x\neq-1$ or $3$)

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