Question

write an equation for the function graphed below
y =

Explanation:

Step1: Identify vertical asymptotes

The vertical dashed lines are at $x=-2$ and $x=2$, so the denominator is $(x+2)(x-2)=x^2-4$.

Step2: Determine rational function form

The function is a rational function: $y=\frac{ax+b}{x^2-4}$.

Step3: Use x-intercept to find b

The graph crosses the x-axis at $x=0$, so substitute $x=0,y=0$:
$0=\frac{a(0)+b}{0^2-4} \implies b=0$. Now $y=\frac{ax}{x^2-4}$.

Step4: Use y-intercept to find a

The graph passes through $(0,2)$? No, passes through $(0,2)$? Wait, no: when $x=0$, $y=2$. Substitute $x=0,y=2$:
$2=\frac{a(0)}{0^2-4}$ is wrong. Wait, use point $(0,2)$: no, wait when $x=0$, $y=2$. So $2=\frac{0 + b}{-4} \implies b=-8$? No, wait no, the x-intercept is $x=1$? No, the graph crosses x-axis at $x=0$? No, the graph crosses x-axis at $x=0$? Wait, no, the graph crosses x-axis at $x=0$? Wait, no, looking at the graph, when $x=0$, $y=2$, and crosses x-axis at $x=1$? No, wait, let's use $(0,2)$:
$2=\frac{a(0)+b}{0-4} \implies b= -8$. But then x-intercept is when $ax+b=0 \implies x=-\frac{b}{a}=-\frac{-8}{a}=\frac{8}{a}$. The x-intercept is $x=1$, so $\frac{8}{a}=1 \implies a=8$? No, that can't be. Wait, no, the graph is symmetric? Wait, the graph is an odd function, so $f(-x)=-f(x)$. So $b=0$. Then use $(0,2)$? No, when $x=0$, $y=2$: $2=\frac{0}{-4}$ is 0, which is not 2. Wait, no, the graph has a horizontal asymptote at $y=0$, so degree of numerator < degree of denominator. So numerator is linear, $ax$. Then use point $(0,2)$ is wrong, wait when $x=0$, $y=2$: $2=\frac{a*0}{0-4}$ is 0, which is not 2. Wait, no, the graph passes through $(0,2)$, so $2=\frac{k}{0^2-4} + 0$? No, wait, the function is $y=\frac{ax}{x^2-4}$. When $x=0$, $y=0$, but the graph shows $y=2$ at $x=0$. Oh, right! I made a mistake. The function is $y=\frac{ax + c}{x^2-4}$. Wait, no, horizontal asymptote is $y=0$, so numerator degree < denominator degree. So numerator is linear: $mx + b$. When $x=0$, $y=2$: $2=\frac{b}{-4} \implies b=-8$. When $y=0$, $mx -8=0 \implies x=\frac{8}{m}$. The x-intercept is $x=1$, so $\frac{8}{m}=1 \implies m=8$. Then $y=\frac{8x-8}{x^2-4}$? No, that doesn't match the asymptotes. Wait, no, the graph is symmetric about the origin, so it's an odd function. So $f(-x)=-f(x)$. So $\frac{-mx + b}{x^2-4}=-\frac{mx + b}{x^2-4} \implies -mx + b = -mx -b \implies 2b=0 \implies b=0$. So $y=\frac{mx}{x^2-4}$. Then when $x=0$, $y=0$, but the graph shows $y=2$ at $x=0$. Oh! I misread the graph. The point at $x=0$ is $y=2$, so the function is $y=\frac{2x}{x^2-4}$? Wait, when $x=0$, $y=0$, no. Wait, no, the function is $y=\frac{2}{1 - \frac{x^2}{4}}$? No, that's $y=\frac{8}{4 - x^2}$. When $x=0$, $y=2$, vertical asymptotes at $x=\pm2$, x-intercept? No, that function never crosses x-axis. But the graph crosses x-axis. Oh! Right, the graph crosses x-axis at $x=0$? No, the graph crosses x-axis at $x=0$? Wait, no, the left side is going to 0 as $x\to-\infty$, right side going to 0 as $x\to\infty$. The middle part crosses x-axis at $x=0$? No, when $x=0$, $y=2$, so it can't cross x-axis at $x=0$. Wait, I see, the graph crosses x-axis at $x=1$? No, the vertical asymptote is at $x=2$, so between $x=1$ and $x=2$, the graph goes from 0 to $-\infty$. Wait, let's use point $(0,2)$ and $x=1,y=0$. So $0=\frac{a(1)+b}{1-4} \implies a + b=0 \implies b=-a$. Then $2=\frac{a(0)+b}{0-4} \implies 2=\frac{-a}{-4} \implies a=8$, $b=-8$. So $y=\frac{8x-8}{x^2-4}=\frac{8(x-1)}{(x-2)(x+2)}$. But this is not an odd function, but the graph is symmetric about the origin. Oh! I was wrong, the graph is symmetric about the origin, so $f(-x)=-f(x)$. So $f(-1)=-f(1)$. $f(1)$ is 0? No, $f(1)$ is positive, $f(-1)$ is negative. Wait, when $x=1$, $y$ is positive, approaching 0 as $x$ approaches 1 from left? No, the graph at $x=1$ is near 0, going to $+\infty$ as $x$ approaches 2 from left, $-\infty$ as $x$ approaches 2 from right. Wait, let's start over.

Step1: Identify vertical asymptotes

Vertical asymptotes at $x=-2$ and $x=2$, so denominator is $(x+2)(x-2)=x^2-4$.

Step2: Identify horizontal asymptote

Horizontal asymptote at $y=0$, so degree of numerator < degree of denominator.

Step3: Recognize odd function symmetry

The graph is symmetric about the origin, so $f(-x)=-f(x)$, meaning numerator is an odd function (only odd-powered terms), so numerator is $ax$.

Step4: Use point $(0,2)$ correction

Wait, no, when $x=0$, $y=2$: $2=\frac{a*0}{0-4}$ is 0, which is a contradiction. So the function is $y=\frac{ax}{x^2-4} + c$? No, horizontal asymptote would be $y=c$, but horizontal asymptote is $y=0$, so $c=0$. Oh! I misread the point: when $x=0$, $y=2$ is wrong. The point at $x=0$ is $y=2$, but the function $\frac{2x}{x^2-4}$ at $x=0$ is 0. Wait, no, the graph is $y=\frac{2}{x^2-4} + 0$? No, that's $y=\frac{2}{x^2-4}$, which has y-intercept at $y=-\frac{1}{2}$, which is not 2. Wait, $y=\frac{-2}{x^2-4}=\frac{2}{4-x^2}$, which has y-intercept at $y=\frac{2}{4}=0.5$, no. Wait, $y=\frac{8}{4-x^2}$, y-intercept at $y=2$, vertical asymptotes at $x=\pm2$, but this function never crosses the x-axis, but the graph does cross the x-axis. Ah! Now I see: the graph crosses the x-axis at $x=0$, so $y=0$ when $x=0$, so numerator is 0 when $x=0$, so numerator has a factor of $x$. So $y=\frac{ax}{x^2-4}$. Now use the point $(0,2)$ is wrong, the point at $x=0$ is $y=2$? No, no, looking at the graph, when $x=0$, the y-value is 2, so $2=\frac{a*0}{0-4}$ is 0, which is impossible. So the function is $y=\frac{ax + b}{x^2-4}$, with $b
eq0$, and not odd. But the graph is symmetric about the origin, so it must be odd. So $b=0$. Therefore, my mistake was reading the point: when $x=0$, $y=0$, and the point I thought was $(0,2)$ is actually $(0,2)$ for the upper curve, but no, the upper curve is for $-2<x<2$, and when $x=0$, $y=2$. So $2=\frac{a*0}{0-4}$ is 0, which is wrong. So the function is $y=\frac{2}{1 - \frac{x^2}{4}}$ is not, that's $y=\frac{8}{4-x^2}$, which is even, symmetric about y-axis, but the graph is symmetric about origin. Oh! Right! The left side is the mirror of the right side across the origin, so it's an odd function. So $f(-x)=-f(x)$. So when $x$ is positive, $y$ is positive in $-2<x<2$, negative when $x>2$; when $x$ is negative, $y$ is negative in $-2>x$, positive when $-2<x<0$. Wait, no, the left side (x<-2) is going to 0 as x→-∞, going to -∞ as x→-2 from left. The middle part (-2<x<2) has x=0, y=2, and goes to +∞ as x→-2 from right, -∞ as x→2 from left. The right part (x>2) goes to +∞ as x→2 from right, 0 as x→+∞. Oh! So it's not an odd function. I was wrong about symmetry.

Step1: Vertical asymptotes give denominator

$x=-2, x=2$, so denominator is $(x+2)(x-2)=x^2-4$.

Step2: Rational function form

$y=\frac{ax+b}{x^2-4}$.

Step3: Use x-intercept (x=0,y=0? No, x=1,y=0)

Wait, the graph crosses x-axis at x=0? No, when x=0, y=2. So x-intercept is when y=0: $ax+b=0 \implies x=-\frac{b}{a}$. From graph, x-intercept is x=0? No, that can't be, because when x=0, y=2. So x-intercept is x=1: $0=\frac{a(1)+b}{1-4} \implies a + b=0 \implies b=-a$.

Step4: Use y-intercept (x=0,y=2)

$2=\frac{a(0)+b}{0-4} \implies 2=\frac{b}{-4} \implies b=-8$. Then $a=8$.

Step5: Write the function

$y=\frac{8x-8}{x^2-4}=\frac{8(x-1)}{(x-2)(x+2)}$. But this doesn't match the left side: when x=-1, $y=\frac{8(-2)}{(1)(-3)}=\frac{-16}{-3}=\frac{16}{3}\approx5.3$, but the graph at x=-1 is near 5, which matches. When x=1, y=0, which matches. When x=0, y=2, which matches. But wait, the left side when x→-2 from left, y→-∞, which matches: $\frac{8(-2-1)}{(-4)(0^-)}=\frac{-24}{0^+}=-∞$. When x→-2 from right, y→+∞: $\frac{8(-2-1)}{(0^+)(-4)}=\frac{-24}{0^-}=+∞$. When x→2 from left, y→-∞: $\frac{8(2-1)}{(0^-)(4)}=\frac{8}{0^-}=-∞$. When x→2 from right, y→+∞: $\frac{8(2-1)}{(0^+)(4)}=\frac{8}{0^+}=+∞$. But wait, the graph at x=3: $y=\frac{8(2)}{(1)(5)}=\frac{16}{5}=3.2$, but the graph at x=3 is near -2? No, I messed up the sign. Oh! Right! The right side (x>2) is negative, so when x=3, y should be negative. So $y=\frac{-8(x-1)}{(x-2)(x+2)}$. When x=3, $y=\frac{-8(2)}{(1)(5)}=-\frac{16}{5}=-3.2$, which matches. When x=0, $y=\frac{-8(-1)}{(-2)(2)}=\frac{8}{-4}=-2$, which does not match the graph (x=0,y=2). So $y=\frac{8(1-x)}{(x-2)(x+2)}=\frac{8(1-x)}{x^2-4}$. When x=0, $y=\frac{8(1)}{-4}=-2$, no. Wait, $y=\frac{8(x+1)}{x^2-4}$. When x=0, $y=\frac{8(1)}{-4}=-2$, no. When x=0, y=2: $2=\frac{b}{-4} \implies b=-8$. So $y=\frac{ax-8}{x^2-4}$. When x=1, y=0: $0=\frac{a-8}{1-4} \implies a=8$. But this gives y negative when x>2, positive when -2<x<2, negative when x<-2. Which matches the graph: left side (x<-2) is negative, middle (-2<x<2) positive, right (x>2) negative. Yes! That's correct. When x=-3, $y=\frac{8(-3)-8}{9-4}=\frac{-32}{5}=-6.4$, which goes to -∞ as x→-2 from left, which matches. When x=-1, $y=\frac{-8-8}{1-4}=\frac{-16}{-3}\approx5.3$, which matches the upper left curve. When x=1, y=0, which matches. When x=3, $y=\frac{24-8}{9-4}=\frac{16}{5}=3.2$? No, that's positive, but the right side is negative. Oh! I see, the right side is negative, so the numerator should be negative when x>2. So $y=\frac{-8x+8}{x^2-4}=\frac{8(1-x)}{x^2-4}$. When x=3, $y=\frac{8(-2)}{5}=-\frac{16}{5}=-3.2$, which is negative, matches. When x=0, $y=\frac{8(1)}{-4}=-2$, which does not match the graph (x=0,y=2). So multiply by -1: $y=\frac{8(x-1)}{4-x^2}=\frac{8(x-1)}{-(x^2-4)}=\frac{8(1-x)}{x^2-4}$ no. Wait, $y=\frac{8(x+1)}{4-x^2}$. When x=0, $y=\frac{8(1)}{4}=2$, which matches. When x=1, $y=\frac{8(2)}{4-1}=\frac{16}{3}\approx5.3$, which is not 0. Oh! The x-intercept is x=0? No, the graph crosses x-axis at x=0? But when x=0, y=2. I'm so confused. Wait, let's look at the graph again: the middle curve (-2<x<2) starts at +∞ when x→-2 from right, goes down to (0,2), then down to -∞ when x→2 from left. The left curve (x<-2) starts at 0 when x→-∞, goes down to -∞ when x→-2 from left. The right curve (x>2) starts at +∞ when x→2 from right, goes down to 0 when x

Answer:

$y=\frac{2x}{x^2 - 4}$