Sketch a function from its key characteristics

Using characteristics such as domain and range, intercepts and zeros, symmetry, discontinuities and asymptotes, end behavior, intervals of increase or decrease, extrema, concavity, and periodicity where applicable, the learner sketches a coherent graph showing how function values change. The representation distinguishes exact features from qualitative behavior and requires checking that the characteristics are mutually consistent, rather than joining plotted points arbitrarily or assuming continuity. This scope excludes highly general, parametric, or advanced implicit-curve sketching.

Detailed Explanation: Sketch a function from its key characteristics

To sketch a function from its characteristics, identify the exact features first, then use the function’s behavior to connect them. Do not connect points across a discontinuity or assume the graph is continuous everywhere.

Worked example

Sketch

f(x)=\frac{x-2}{x+1}.

1. Find the domain and vertical asymptote

The denominator cannot equal zero:

x+1\neq 0 \quad\Rightarrow\quad x\neq -1.

So the domain is

\boxed{x\neq -1}.

Because the denominator is zero at $x=-1$ , there is a vertical asymptote:

\boxed{x=-1}.

The graph cannot cross or connect through this vertical line.

2. Find the intercepts

For the $x$ -intercept, set $f(x)=0$ . A fraction is zero when its numerator is zero:

x-2=0 \quad\Rightarrow\quad x=2.

So the $x$ -intercept is

\boxed{(2,0)}.

For the $y$ -intercept, substitute $x=0$ :

f(0)=\frac{0-2}{0+1}=-2.

So the $y$ -intercept is

\boxed{(0,-2)}.

3. Find the horizontal asymptote and end behavior

Rewrite the function:

f(x)=\frac{x-2}{x+1}=1-\frac{3}{x+1}.

As $x$ becomes very large in either direction, $\frac{3}{x+1}$ approaches $0$ . Therefore,

f(x)\to 1 \quad\text{as }x\to\infty\text{ or }x\to-\infty.

The horizontal asymptote is

\boxed{y=1}.

The graph approaches this line but never reaches it. In fact, solving $f(x)=1$ gives

1-\frac{3}{x+1}=1,

which is impossible. Thus the range is

\boxed{y\neq 1}.

4. Determine the behavior near the vertical asymptote

Use

f(x)=1-\frac{3}{x+1}.

As $x\to -1^-$ , $x+1$ is a small negative number, so $-\frac{3}{x+1}$ is very large and positive. Thus

f(x)\to+\infty.

As $x\to -1^+$ , $x+1$ is a small positive number, so $-\frac{3}{x+1}$ is very large and negative. Thus

f(x)\to-\infty.

This tells us how the two separate branches behave near $x=-1$ .

5. Determine whether each branch increases or decreases

Differentiate:

f'(x)=\frac{3}{(x+1)^2}.

Since $(x+1)^2>0$ whenever $x\neq -1$ ,

f'(x)>0.

Therefore, the function is increasing on both intervals:

\boxed{(-\infty,-1)} \qquad\text{and}\qquad \boxed{(-1,\infty)}.

6. Plot a few points and sketch

Useful points include

f(-2)=4,\qquad f(0)=-2,\qquad f(2)=0.

Now assemble the information:

Draw dashed asymptotes $x=-1$ and $y=1$ .
On the left of $x=-1$ , the branch increases from just above $y=1$ to $+\infty$ .
On the right of $x=-1$ , the branch increases from $-\infty$ toward $y=1$ from below.
Plot $(-2,4)$ on the left branch.
Plot $(0,-2)$ and $(2,0)$ on the right branch.
Do not join the two branches across $x=-1$ .

A rough sketch has this shape:

          y
          ↑
      left branch       y = 1  - - - - - - - - - - -
             /                         right branch
            /                              ______
           /                         _____/
          /                    _____/
---------|--------------------|--------------------→ x
        x=-1                 (0,-2)       (2,0)
          \ 
           \ 
            ↓

The important features are the two asymptotes, the two separate branches, the intercepts, and the correct end behavior. These characteristics are mutually consistent with the function’s domain and range.

Learn by doing: Sketch a function from its key characteristics

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Sinusoidal Function Parameters (4 Params) - Function to Graph

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