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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

Kortlek
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Harrison
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    Is the assignment to write the equation of any function that satisfies these characteristics? Are you allowed to write a piecewise equation (i.e. $f(x)$ when $a < x < b$, but $g(x)$ otherwise)? – GMB Jul 02 '13 at 03:55
  • Yes, but I am asked for an exact equation so piecewise would not work unfortunately. (any equation that satisfies these characteristics) – Harrison Jul 02 '13 at 04:07
  • Equations are not functions, and a function defined piecewise is still 'exact'. – anomaly Jun 26 '15 at 16:32

3 Answers3

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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.

Ross Millikan
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  • Hi, thanks for your help. I know how to do every step of this problem, but I was caught with the 'no x intercepts' part; I forgot to tell you that the answer to this graph isn't supposed to be linear. – Harrison Jul 02 '13 at 04:05
  • You can multiply it by a quadratic factor. If you pick a quadratic that is always positive and has value $1$ at $x=-1$ and $x=3$ you will be there. – Ross Millikan Jul 02 '13 at 04:08
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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 $3$)

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

Maazul
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