Exercises — Vertical line test for functions

Panel A (slanted line): Slide a vertical line anywhere. At every position it touches the slanted line in exactly one spot — the line is never "stacked" above itself. So $0$ or $1$ hits everywhere → passes → it is a function.

Panel B (circle): Put the vertical line through the middle of the circle. Look at the two red dots — it hits the top of the circle and the bottom of the circle: that is two points at once. One input $x$, two outputs $y$ → fails → not a function.

Why the circle fails and the line doesn't: the line never doubles back on itself in the $y$-direction; the circle bulges above and below the same $x$.

The trick: if the equation is already solved as "$y=(\text{something in }x)$" with no $\pm$ and no even power on $y$, then plugging in one $x$ produces exactly one $y$.

• (a) $y=3x-1$: one output. Function.
• (b) $y=|x|$: absolute value gives a single non-negative number. Function.
• (d) $y=2^x$: exponent gives one value. Function.
• (c) $y^2=x$: here $y$ is squared, so solving gives $y=\pm \sqrt{x}$ — the $\pm$ is the warning sign of two outputs. Not a function.

Answer: (a), (b), (d) are functions; (c) is not.

Substitute $x=3$ to fix the input, then solve for the output(s): $3^2+y^2=25⟹9+y^2=25⟹y^2=16⟹y=\pm 4$ What this means: the single input $x=3$ produced two outputs, $y=4$ and $y=-4$. The vertical line $x=3$ hits the graph at $(3,4)$ and $(3,-4)$. Conclusion: two hits → not a function. ✗

The symbol  ​ means the principal (non-negative) root — by definition it returns just one number, never $\pm$.

• $x=5$: $y=\sqrt{5-1}=\sqrt{4}=2$. Exactly one output. One hit.
• $x=0$: $\sqrt{0-1}=\sqrt{-1}$ is not a real number, so $x=0$ has no output. The vertical line $x=0$ misses the graph — zero hits.

Every allowed $x$ gives exactly one $y$, and $x$'s outside the domain give none. Never two. Passes → it is a function. ✓

Note: zero hits (like at $x=0$) is fine — it just means $x=0$ is not in the domain. Only $\ge 2$ hits breaks the rule.

$|y|=x$ means "the distance of $y$ from zero equals $x$." At $x=7$: $|y|=7⟹y=7\text{or}y=-7$ Two outputs for one input → two hits at $(7,7)$ and $(7,-7)$ → not a function. ✗

(This is really a sideways "V", opening rightward — it bulges above and below the axis exactly like the circle did.)

A piecewise rule only ever uses one line of the definition for any given $x$ — the two cases don't overlap.

• $x=1$ satisfies "$x\ge 1$", so we use $5-x$: $f(1)=5-1=4$. The first line ("$x<1$") does not apply at $x=1$ — that inequality is strict. So $x=1$ has the single output $4$.
• Any other $x$ falls into exactly one case → exactly one output.

No $x$ ever gets two outputs → passes → function. ✓

Even if there is a visible jump at $x=1$ (the top piece approaches $1+2=3$ while the bottom starts at $4$), a jump is a change in output, not two outputs at the same $x$.

An open circle (hollow) means "this point is not included"; a filled dot means "this point is included." At $x=2$ the vertical line passes through the open circle at $(2,1)$ — but that point is excluded, so it doesn't count as a hit — and through the filled dot at $(2,4)$, which is a hit. So the vertical line $x=2$ actually strikes the graph at exactly one genuine point, $(2,4)$. Passes → function. ✓

Key idea: open circles are "ghost" points — the vertical line seems to hit two dots, but only the solid ones count.

Solve for $y$: $y=\pm \sqrt{x-3}$.

• Two outputs whenever $\sqrt{x-3}$ is a positive number, i.e. when $x-3>0\Rightarrow x>3$. Every such $x$ gives $y=+\sqrt{x-3}$ and $y=-\sqrt{x-3}$ → fails.
• Exactly one output at $x=3$: then $y=\pm \sqrt{0}=0$, and $+0$ and $-0$ are the same number. One point $(3,0)$.
• No points when $x-3<0\Rightarrow x<3$: the square root of a negative isn't real.

Because there exist $x$'s (all $x>3$) with two outputs, the whole relation is not a function. The single-point tip at $x=3$ doesn't rescue it.

The circle fails because each $x$ (strictly between $-5$ and $5$) gives one output above and one below the axis. Keep only the top half: $y=+\sqrt{25-x^2}$ Now each $x$ gives a single non-negative $y$ → passes.

• Domain: we need $25-x^2\ge 0\Rightarrow -5\le x\le 5$, i.e. $[-5,5]$ (see 2.3.02-Domain-and-rangefrom-graphs).
• Range: $y$ runs from $\sqrt{25-25}=0$ up to $\sqrt{25-0}=5$, i.e. $[0,5]$.

(Choosing the bottom half $y=-\sqrt{25-x^2}$ also works; its range would be $[-5,0]$.)

• Vertical line test: each $x$ gives one $y=x^2$. Passes → it is a function. ✓
• Horizontal line test (from 2.2.05-Horizontal-line-test-for-injective-functions): the line $y=4$ hits the parabola at $x=2$ and $x=-2$ — two hits → fails → the function is not one-to-one.

Why it matters: to have an inverse function you must pass both tests. $y=x^2$ passes vertical but fails horizontal, so it has no inverse unless you restrict its domain (say to $x\ge 0$) to make it one-to-one. Vertical = "is a function at all"; horizontal = "can it be undone."

• $x=4$ (vertical line): the vertical test line $x=4$ lies on top of it and hits it at infinitely many points (every $y$). Two or more hits → not a function. This is the most extreme failure: one input, endless outputs.
• $y=4$ (horizontal line): any vertical line hits it in exactly one point $(a,4)$. Passes → it is a function (the constant function $f(x)=4$). Every input maps to the single output $4$.

So "line" alone tells you nothing — direction is everything.

One clean answer: $f(x)=\{\begin{array}{ll}-1& x<0\\ +1& x>0\end{array}$

• Function: every $x\ne 0$ lands in exactly one case → one output. Passes vertical test. ✓
• Not continuous: as $x$ crosses $0$ the output jumps from $-1$ to $+1$.
• Undefined at $x=0$: neither case includes $x=0$, so the vertical line $x=0$ hits nothing (zero crossings — allowed).

This shows a function may jump and have gaps and still be perfectly legal, echoing Mistake 2 of the parent note.

False. Passing the horizontal test only means no output is repeated — it says nothing about inputs. Counterexample: the sideways parabola $x=y^2$.

• Horizontal line $y=b$ hits it once (at $x=b^2$) → passes horizontal.
• But vertical line $x=4$ hits it at $y=2$ and $y=-2$ → fails vertical → not a function.

The horizontal test can never certify functionhood; only the vertical test can.

Scan the inputs: $x=-1$ appears once; $x=2$ appears twice, paired with $y=5$ and $y=7$. The vertical line $x=2$ hits two points → not a function. ✗

Repair: delete either $(2,5)$ or $(2,7)$ so that $x=2$ has a single output. Deleting $(2,7)$ leaves $\{(-1,0),(2,5)\}$:

• Domain = set of used inputs = $\{-1,2\}$.
• Range = set of used outputs = $\{0,5\}$.

Now each input has exactly one output → function. ✓