The graph of an equation in \(x\) and \(y\) is the set of all points with coordinates \((x,y)\) satisfying the equation. In the case of a function \(y = f ( x )\), the graph of \(f\) is the point set \(\{ (x,y)\, | \,y = f ( x ) \}\).

The most fundamental way to sketch the graph of a function is to plot points.

- Select an interval on the \(x\)-axis encompassing all or part of the domain of the function.
- Select a finite set of \(x\) values in the domain of the function lying in that selected interval.
- For each of the selected values of \(x\), compute the corresponding value of \(y\) using the formula for \(y = f ( x )\).
- Plot this finite set of points \((x,y)\) in the plane and sketch the graph by connecting the points, being careful not to connect two adjacent points unless all the input values lying between the two points are in the domain of the function.

There are certain shortcuts which may make the task of sketching graphs easier by reducing the number of points which must be plotted. These shortcuts are

- symmetry
- shifting, i.e. translation
- stretching or shrinking
- reflecting

These shortcuts apply **not only** to the graphs of functions (vertical line test) but to **all** graphs having an equation in \(x\) and \(y\).

Let \(f(x)=4-x^2\) on the interval \([ - 3, 3 ]\). Replace the \(x\) in the equation with \(( - x )\) and simplify. Did you get the same equation? Sketch the graph and notice that for every point \(( x, y )\) on the graph, the mirror image of that point in the \(y\)-axis, namely the point \(( - x, y )\), is also on the graph.

See Solution
A function which is symmetric with respect to the \(y\)-axis is called an **even** function and has the special property that \(f ( - x ) = f ( x )\) for every \(x\) in the domain of the function.

A graph is symmetric with respect to the \(x\)-axis if the point \(( x, - y )\) is on the graph whenever the point \(( x, y )\) is on the graph.

This means that if one replaces every occurrence of \(y\) in the equation with \( - y )\) [ be sure to put parentheses around the \(- y\) ] and then simplifies the equation, the resulting equation will be equivalent to the original equation.

Note that if \(y\) is not 0 then \(( x, y )\) and \(( x, - y )\) are two points which lie on the same vertical line, so the graph would not be the graph of a function in that case. [ Why not? ]

Let \(x=y^2-1\) for \(y\) in the interval \([ - 2, 2 ]\). Replace the \(y\) in the equation with \(( - y )\) and simplify. Did you get the same equation? Sketch the graph and notice that for every point \(( x, y )\) on the graph, the mirror image of that point in the \(x\)-axis, namely the point \(( x, - y )\), is also on the graph.

See SolutionA graph is symmetric with respect to the origin if the point \(( - x, - y )\) is on the graph whenever the point \(( x, y )\) is on the graph.

This means that if one replaces every occurrence of \(x\) in the equation with \(( - x )\) and every occurrence of \(y\) in the equation with \(( - y )\) [ be sure to put parentheses around the \(- x\) and \(- y\) ] and then simplifies the equation, the resulting equation will be equivalent to the original equation.

Notice that the point \(( - x, - y )\) is the same distance from the origin as the point \(( x, y )\) , except it is on the opposite side of the origin and that the line connecting the points \(( x, y )\) and \(( - x, - y )\) always passes through the origin \((0,0)\).

Let \(y=x^3\) for x in the interval \([ - 2, 2 ]\). Replace the \(x\) in the equation with \(( - x )\) and replace the \(y\) in the equation with \(( - y )\) and simplify. Did you get the same equation? Sketch the graph and notice that for every point \(( x, y )\) on the graph, the point \(( - x, - y )\), is also on the graph.

See Solution
A function which is symmetric with respect to the origin is called an **odd** function and has the special property that \(f ( - x ) = - f ( x )\) for every \(x\) in the domain of the function.

In mathematics, to **translate** a geometrical figure means to change it's position without distorting or rotating it.

The **horizontal translation principle** states that replacing each occurrence of \(x\) in an \(x y\) equation with
\(( x - h )\) [ be sure to put parentheses around the \(x - h\) ] produces an equation whose graph lies \(h\) units horizontally from the graph of the original equation. The \(h\) can be any real number, positive or negative.

Sketch the graph of \(y=x^2\), \(y=(x-1)^2\) and \(y=(x+1)^2\). Notice the application of the horizontal translation principle.

See SolutionStated in terms of function notation, the horizontal translation principle states that the graph of \(y = f ( x - h )\) is shifted \(h\) units horizontally from the graph of \(y = f ( x )\).

Thus the graph of \(y = f ( x - 2 )\) will lie two units to the right of the graph of \(y = f ( x )\) and the graph of \(y = f ( x + 3 ) = f ( x - ( - 3 ) )\) will lie three units to the left of y = f ( x ). That is, **minus** \(3\) units means three units to the left.

Sketch the graph of \(f(x)=\sqrt{x}\). Then sketch the graphs of \(g(x)=f(x-2)=\sqrt{x-2}\) and \(p(x)=f(x+1)=\sqrt{x+1}\).

See Solution
The **vertical translation principle** states that replacing each occurrence of \(y\) in an \(x y\) equation with \(( y - k )\) [ be sure to put parentheses around the \(y - k\) ] produces an equation whose graph lies \(k\) units vertically from the graph of the original equation. The \(k\) can be any real number, positive or negative.

Sketch the graph of \(y=x^2\), \(y-1=x^2\) and \(y+1=x^2\). Notice the application of the vertical translation principle.

See SolutionStated in terms of function notation, the vertical translation principle states that the graph of \(y - k = f ( x )\) [ alternately, \(y = f ( x ) + k\) ] is shifted \(k\) units vertically from the graph of \(y = f ( x )\).

Thus the graph of \(y - 2 = f ( x )\) will lie two units above the graph of \(y = f ( x )\) and the graph of \(y + 3= f ( x )\) will lie three units below \(y = f ( x )\).

Sketch the graph of \(f(x)=\sqrt{x}\).Then sketch the graphs of \(g(x)=\sqrt{x}+2\) and \(p(x)=\sqrt{x}-1\).

See Solution

Both the horizontal and vertical translation principles may be applied to the same problem.

Given an equation in \(x\) and \(y\) if each occurrence of \(x\) is replaced by \((x-h)\) and each occurrence of \(y\) is replaced by \((y-k)\) then the resulting equation will have a graph equivalent to shifting the graph of the original equation \(h\) units horizontally and \(k\) units vertically. Stated in function notation, if the original function is \(y=g(x)\) then the equation of the translated function will be \(y-k=g(x-h)\). Solving for \(y\) gives \(y=g(x-h)+k\) as the equation for the new function. If we call the new function \(f\) then we have \(f(x)=g(x-h)+k\). So the graph of \(f\) is the graph of \(g\) shifted \(h\) units horizontally and \(k\) units vertically.

Sketch the graph of \(f(x)=\sqrt{x-2}+1\) using both the horizontal and vertical translation principles.

See Solution
The **horizontal stretching/shrinking principle** states that replacing each occurrence of \(x\) in an \(x y\) equation with \(\left(\dfrac{x}{c}\right)\) **stretches** the graph horizontally by a factor of \(c\) when \(c > 1\) and **shrinks** the graph horizontally by a factor \(c\) when \(0 < c < 1\).

Sketch the graph of \(y=\left(\dfrac{x}{2}\right)^2\) together with the graph of \(y=x^2\). Notice that the first graph is **twice as wide** as the second.

Sketch the graph of \(y=(2x)^2=\left(\dfrac{x}{1/2}\right)^2\) together with the graph of \(y=x^2\). Notice that the first graph is only **one-half** as wide as the second.

The vertical stretching/shrinking principle states that replacing each occurrence of \(y\) in an \(x y\) equation with \(\left(\dfrac{y}{c}\right)\) **stretches** the graph vertically by a factor of \(c\) when \(c > 1\) and **shrinks** the graph vertically by a factor \(c\) when \(0 < c < 1\).

Sketch the graph of \(\dfrac{y}{2}=x^2\) [ after first re-writing as \(y=2x^2\) ] together with the graph of \(y=x^2\). Notice that the first graph is **twice as tall** as the second.

Sketch the graph of \(f(x)=4x^2\) together with the graph of \(g(x)=x^2\). Compare this problem to exercise 1.7.12.

Notice that this particular problem can be worked either by vertical stretching or by horizontal shrinking!

See Solution
Sketch the graph of \(y=\dfrac{x^2}{2}\) [ that is, \(2y=x^2\) or \(\dfrac{y}{1/2}=x^2\) ], together with the graph of \(y=x^2\). Notice that the first graph is only **one-half** as tall as the second.

Sketch the graph of \(f(x)=\left(\dfrac{x}{\sqrt{2}}\right)^2\) together with the graph of \(g(x)=x^2\). Hint: Compare to exercise 1.7.15. Notice that the graph of \(f(x)=\left(\dfrac{x}{\sqrt{2}}\right)^2\) is wider than the graph of \(g(x)=x^2\).

See Solution
The **vertical reflection principle** states that replacing each occurrence of \(x\) in an \(x y\) equation with \(( - x )\) [ be sure to put parentheses around the \(- x\) ] reflects the graph of the original equation about the \(\mathbf{y}\)**-axis**.

The **horizontal reflection principle** states that replacing each occurrence of \(y\) in an \(x y\) equation with
\(( - y )\) [ be sure to put parentheses around the \(- y\) ] reflects the graph of the original equation about the \(\mathbf{x}\)**-axis**.

Sketch the graphs of \(y=\sqrt{x}\) and \(-y=\sqrt{x}\) [ which is equivalent to \(y=-\sqrt{x}\) ]. Note how this example differs from Exercise 1.7.17 above.

See SolutionSketch the graph of \(f(x)=-\sqrt{1-x}\) using the principles of reflection and translation.

Hint: First, graph \(-y=\sqrt{-x}\) and re-write the equation for \(f(x)\) in the equivalent form \(-y=\sqrt{-(x-1)}\)

See Solution