Graphs of Functions: Practice

Substitute $-x$ and use that even powers erase the sign.

$f(-x) = (-x)^4 - (-x)^2 = x^4 - x^2 = f(x)$

Since $f(-x) = f(x)$, the function is even. Both terms have even degree, which is why the parity survives.

The denominator is zero at $x = 3$, so $3$ is excluded. Factor and cancel for the other inputs.

$\frac{(x - 3)(x + 2)}{x - 3} = x + 2, \qquad x \ne 3$

Evaluate the simplified rule at the excluded input: $3 + 2 = 5$, so the hole is at $(3, 5)$.

Factor out the highest power to expose the far-out behavior.

$f(x) = x^3\left(1 - \frac{4}{x^2}\right)$

For $\lvert x \rvert > 2$ the bracket is positive, so $f$ has the sign of $x^3$. For large negative $x$, $x^3$ is negative and large, so the graph falls to large negative values.

Evaluate $g$ at $x = 0$ and set it equal to $-6$.

$g(0) = 0 + c(-1)(-2)(-3) = -6c = -6$

Solving $-6c = -6$ gives $c = 1$.

Factor out $x$, then factor the difference of squares.

$x^3 - 9x = x(x^2 - 9) = x(x - 3)(x + 3) = 0$

The zeros are $x = 0$, $x = 3$, and $x = -3$.

The condition $f(-x) = f(x)$ is the definition of an even function.

$f \text{ even} \quad \Longleftrightarrow \quad \text{graph unchanged by reflection across the y-axis}$

Reflection sends $(x, f(x))$ to $(-x, f(x)) = (-x, f(-x))$, which is already a point of the graph, so the picture is unchanged.

Test $x^2 + x$ by substituting $-x$.

$f(-x) = (-x)^2 + (-x) = x^2 - x$

This equals neither $f(x) = x^2 + x$ nor $-f(x) = -x^2 - x$ (except at $x = 0$), so it is neither. The others are even, odd, and even in turn.

Crossing versus touching is decided by the power of each factor.

$(x - 1)^2 \text{ even power} \Rightarrow \text{touch at } 1, \qquad (x - 4) \text{ odd power} \Rightarrow \text{cross at } 4$

So the graph crosses only at $x = 4$, and merely touches at $x = 1$.

Factor out the highest power.

$f(x) = x^4\left(1 - \frac{5}{x^2} + \frac{4}{x^4}\right)$

For $\lvert x \rvert \ge 3$ the bracket is at least $1 - \tfrac{5}{9} = \tfrac{4}{9} > 0$, and $x^4 > 0$, so $f$ rises to large positive values like $x^4$.

The points fit $x^2$, but they fit endless other rules too. For any $c$,

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

agrees at all four inputs, since the product vanishes at each. Different $c$ give different functions, so infinitely many pass through the four points.

The rule is undefined where the denominator is zero.

$x + 2 = 0 \Rightarrow x = -2$

So $x = -2$ is excluded from the domain.

Factor the quadratic and set each factor to zero.

$x^2 - 5x + 6 = (x - 2)(x - 3) = 0 \Rightarrow x = 2 \text{ or } x = 3$

The x-intercepts are at $x = 2$ and $x = 3$.