Graphs of Functions: Additional Practice (Set 2)

Check the multiplicity at each zero. The factor $(x - 2)^2$ has an even power, so the graph touches at $x = 2$; the simple factors $(x + 1)$ and $(x - 4)$ have odd powers, so the graph crosses.

$x = 2 \text{ (touch)}, \quad x = -1 \text{ (cross)}, \quad x = 4 \text{ (cross)}$

The crossings happen only at $x = -1$ and $x = 4$, so there are $2$.

For $x > 3$, all three factors are positive.

$(+)(+)(+) = +$

The product is positive, so the graph is above the axis there.

Subtract $f$ and evaluate the leftover product at $x = 0$.

$g(0) - f(0) = c(0 - 1)(0 - 2) = c(-1)(-2) = 2c$

Set $2c = 4$, giving $c = 2$.

The denominator is zero at $x = 1$, so $1$ is excluded. Factor and cancel.

$\frac{(x + 2)(x - 1)}{x - 1} = x + 2, \qquad x \ne 1$

Evaluate the simplified rule at $x = 1$: $1 + 2 = 3$, so the hole is at $(1, 3)$.

For an odd function, negating the input negates the output.

$f(-3) = -f(3) = -7 \Rightarrow (-3, -7)$

Odd symmetry rotates the point $180^\circ$ about the origin.

Both factors are squared, so both zeros have even multiplicity.

$f(x) = x^2(x + 1)^2 \ge 0 \text{ for all } x$

The sign never changes, so the graph touches the axis at $x = 0$ and $x = -1$ and never crosses; it stays on or above the axis.

Build the sign pattern across the roots $-2, 0, 2$ by multiplying factor signs.

$x < -2:\ -, \quad -2 < x < 0:\ +, \quad 0 < x < 2:\ -, \quad x > 2:\ +$

Two of the four intervals are positive: $-2 < x < 0$ and $x > 2$.

Find the real zeros. The factor $x^2 + 1$ is never zero for real $x$.

$x^2 + 1 > 0, \qquad (x - 2)(x + 2) = 0 \Rightarrow x = \pm 2$

Both real zeros are simple, so the graph crosses at $x = 2$ and $x = -2$, which is $2$ crossings.

Factor out the highest power.

$f(x) = x^3\left(-2 + \frac{1}{x^2}\right)$

For $x > 1$ we have $\tfrac{1}{x^2} < 1$, so the bracket is less than $-1$, hence negative. With $x^3 > 0$ and a negative bracket, $f$ falls to large negative values.

The added product must vanish at every sampled input, which needs a factor for each root, including $x = 0$. The rule missing the factor $x$ fails there.

$1 + (0 - 1)(0 - 2)(0 - 3) = 1 - 6 = -5 \ne 1$

So the rule with product $(x - 1)(x - 2)(x - 3)$ is not consistent; the others all give $1$ at every listed input.

The constant $3$ works, but so does a whole family. For any $c$,

$g(x) = 3 + c\,x(x - 1)(x - 2)$

takes the value $3$ at $x = 0, 1, 2$, since the product vanishes there. Different $c$ give different functions, so infinitely many fit.

The factor $x^2$ has an even power, while $(x - 2)$ and $(x + 2)$ have odd powers.

$x^2 \text{ even} \Rightarrow \text{touch at } 0, \qquad (x \pm 2) \text{ odd} \Rightarrow \text{cross at } \pm 2$

The graph touches only at $x = 0$.