Graphs of Functions: Practice

The solutions of $f(x) = -3$ are where the horizontal line $y = -3$ meets the curve.

$y = -3 \text{ meets the parabola at } x = 0 \text{ and } x = 2$

Two crossings mean two solutions.

The solutions of $f(x) = 0$ are the x-intercepts, where the curve meets the horizontal axis.

$x^2 - 2x - 3 = (x - 3)(x + 1) = 0 \;\Rightarrow\; x = -1 \text{ or } x = 3$

The graph crosses at $(-1, 0)$ and $(3, 0)$.

A turning point is where the direction changes. The direction changes at $x = -1$ (rise to fall) and at $x = 2$ (fall to rise).

$\text{changes at } x = -1 \text{ and } x = 2 \;\Rightarrow\; 2 \text{ turning points}$

The first is a maximum and the second a minimum.

The solutions of $f(x) = 0$ are where the curve meets the horizontal axis.

$\text{crossings at } x = -1 \text{ and } x = 3$

So $f(x) = 0$ has the two solutions $-1$ and $3$.

The slope is the change in output divided by the change in input between the two points.

$\frac{11 - 2}{4 - 1} = \frac{9}{3} = 3$

That slope measures how steeply the graph climbs between the two points.

An x-intercept needs $f(x) = 0$. Count the solutions for each.

$x^2 - 2x + 1 = (x - 1)^2 = 0 \;\Rightarrow\; x = 1 \text{ (once)}$

The others give two intercepts ($\pm 2$, $\pm 3$) or none ($x^2 + 4 > 0$).

The parabola $y = x^2$ falls to the origin then rises.

$\text{decreasing for } x < 0, \quad \text{increasing for } x > 0$

Its single turning point (a minimum) is at the origin, which is also its only x-intercept.

The y-intercept gives the output at $0$, and the x-intercept gives an input with output $0$.

$f(0) = 6 \quad \text{and} \quad f(3) = 0$

So $f(0) = 6$ and $x = 3$ solves $f(x) = 0$.

Crossings of the horizontal axis are the solutions of $f(x) = 0$, and there are none.

$\text{no x-intercepts} \;\Rightarrow\; f(x) = 0 \text{ has no solution}$

The output $0$ is never produced.

Start at $x = -1$ on the horizontal axis and move to the line.

$f(-1) = -3$

The line passes through $(-1, -3)$, so the output there is $-3$.

An upward parabola has its lowest output at the vertex and rises without bound.

$\text{minimum output } -5 \;\Rightarrow\; y \ge -5$

The $-2$ is the input at the vertex, not an output.

A function assigns each input one output, so no two points may share a first coordinate with different second coordinates.

$(2, 0) \text{ and } (2, 3) \text{ share input } 2 \text{ with outputs } 0 \ne 3$

That violates the definition. Sharing an output, as in the other pairs, is perfectly fine.