Question

write an equation for the function graphed below
$y = $
question help: video read written example
submit question

Explanation:

Step1: Identify asymptotes and hole

Vertical asymptotes: $x=-3$, $x=4$; Horizontal asymptote: $y=-1$; Hole at $x=0$ (graph approaches 0 near x=0 but doesn't cross, and the shape matches a rational function with a hole here).

Step2: Set base rational form

General form: $y = -1 + \frac{ax+b}{(x+3)(x-4)x}$ (horizontal asymptote $y=-1$, vertical asymptotes/hole in denominator).

Step3: Use hole behavior at $x\to0$

As $x\to0$, the function approaches 0, so $\frac{b}{0}$ must approach 0, meaning $b=0$. Simplify to $y = -1 + \frac{ax}{x(x+3)(x-4)} = -1 + \frac{a}{(x+3)(x-4)}$.

Step4: Solve for $a$ using a point

Use $x=1$, $y\approx0$:
$0 = -1 + \frac{a}{(1+3)(1-4)}$
$1 = \frac{a}{4(-3)}$
$a = -12$

Step5: Simplify the equation

Substitute $a=-12$:
$y = -1 + \frac{-12}{(x+3)(x-4)}$
$y = \frac{-(x+3)(x-4) -12}{(x+3)(x-4)}$
Expand numerator: $-(x^2 -x -12) -12 = -x^2 +x +12 -12 = -x^2 +x$
So $y = \frac{-x^2 +x}{(x+3)(x-4)}$ or factored $y = \frac{-x(x-1)}{(x+3)(x-4)}$

Answer:

$\boldsymbol{y = -1 - \frac{12}{(x+3)(x-4)}}$ or $\boldsymbol{y = \frac{-x^2 + x}{(x+3)(x-4)}}$