Category Archives: Graphing

Graphing an Exponential Equation by Transformations

Example: For the function below.  Graph using transformations.  Find the y-intercept.  State the horizontal asymptote and the domain and range.

f(x)=-2^{x+2}+2

First we must examine the base function y=2^x

Graph using plotting points.  We can use the standard set of x-values to find ordered pairs.

xy
-22^(-2)=1/4
-12^(-1)=1/2
02^0=1
12^1=2
22^2=4

The graph below shows the points plotted and the line that connects them.  This graph has a horizontal asymptote at y=0.  The domain is (- infty, infty)  and the range is (0, infty)

 

Analyze the transformations.

f(x)=-2^{x+2}+2

The +2 in the exponent shifts the graph left 2 units.

The – in the front of the base reflects the graph over the x-axis.

The +2 next to the base shifts the graph and the horizontal asymptote up two units.

You can see the graph after the transformations.

The horizontal asymptote is y=2.  The domain is (- infty, infty)  and the range is (- infty, 2)

To find the y-intercept we let x=0.

f(0)=-2^{0+2}+2

f(0)=-(2^{2})+2

f(0)=-(4)+2

f(0)=-4+2

f(0)=-2

Thus the y-intercept is (0,-2)

 

 

Quadratic Function

Example: Rewrite the given quadratic function in standard form by completing the square.  Then state the domain, range, vertex, x-intercepts, y-intercept, the orientation (opens up or opens down), and the axis of symmetry.  Finally graph the function.

f(x)=-x^2-2x+8

Solution:

Rewrite function in standard form.

f(x)=-x^2-2x+8

f(x)=-(x^2+2x)+8

f(x)=-(x^2+2x+1)+8+1

f(x)=-(x+1)^2+9

Here is a youTube video that demonstrations the process.

Find the vertex.

When the quadratic function is written in standard form you can identify the vertex as (h,k).

f(x)=a(x-h)^2+k

f(x)=-(x+1)^2+9

The vertex of the quadratic function is (-1,9).

Find the orientation.

The leading coefficient of the quadratic function is negative so the parabola opens down.

Find the y-intercept.

To find a y-intercept let x=0.

f(0)=-(0+1)^2+9

f(0)=-(1)^2+9

f(0)=-1+9

f(0)=8

The y-intercept is (0,8).

Find the x-intercept.

To find an x-intercept let y=0 or f(x)=0.

0=-(x+1)^2+9

0-9=-(x+1)^2+9-9

-9=-(x+1)^2

{-9}/{-1}={-(x+1)^2}/{-1}

9=(x+1)^2

sqrt{9}=sqrt{(x+1)^2}

pm 3=x+1

-1 pm 3=x+1-1

-1 pm 3=x

-1 + 3=x or -1 - 3=x

2=x or -4=x

Find the axis of symmetry.

The axis of symmetry is a vertical line that passes through the vertex.  Since the vertex is (-1,9) the axis of symmetry is x=-1.

Graph the function.

Plot the intercepts and the vertex.

Since the vertex is the highest point we can draw the parabola using the peak at the vertex.

Find the domain and range.

The domain is (- infty, infty).

The range is (  - infty, 9 {}{]}.

Finding the Midpoint

The Midpoint Formula

Suppose A is ( x_1,   y_1) and B is ( x_2,    y_2)

The midpoint of the line segment joining points A and B is given by the following formula.

( {x_1+x_2}/2,   {y_1+y_2}/2)

Example: Find the midpoint of a line segment joining points A and B.

Point A is ( 4, -2) and point B is ( -8, 3)

The x coordinate of the midpoint can be found by averaging the x-values of point A and B.  The y-coordinate of the midpoint can be found by averaging the y-values of point A and B.

( {4+(-8)}/2,   {-2+3}/2)

( {-4}/2,   {1}/2)

( -2,   {1}/2)

X-intercepts and Y-intercepts

An x-intercept is where the graph touches or crosses the x-axis.

A y-intercept is where the graph touches of crosses the y-axis.

In this picture, the graph crosses the x-axis at the ordered pair (2, 0).  Since every ordered pair on the x-axis has a y coordinate of zero we can let y=0 to find x-intercepts.

To find an x-intercept: Let y=0 and solve for x.

In this picture, the graph crosses the y-axis at the ordered pair (0, 6).  Since every ordered pair on the y-axis has a x coordinate of zero we can let x=0 to find y-intercepts.

To find an y-intercept: Let x=0 and solve for y.

Graphing Linear Equations by Finding Intercepts

Steps for Graphing with the Intercept Method

  1. Find the x intercept and the y-intercept.
    • To find an x-intercept let y=0 and solve for x.
    • To find a y-intercept let x=0 and solve for y.
  2. Plot the x-intercept and y-intercept.
  3. Draw the line that connects the intercepts.

Example:   Graph the linear equation 2x-3y = 12

Solution:

 1. Find the x-intercept and the y-intercept.

To find an x-intercept: Let y=0 and solve for x.

2x-3y = 12

2x-3(0) = 12

2x-0 = 12

2x = 12

{2x}/2 = {12}/2

x = 6

The x-intercept of this equation is (6,0)

To find a y-intercept: Let x=0 and solve for y.

2x-3y = 12

2(0)-3y = 12

0-3y = 12

-3y = 12

{-3y}/{-3} = {12}/{-3}

y = -4

The y-intercept of this equation is (0,-4)

2. Plot the x-intercept and the y-intercept.

3. Draw the line that connects the intercepts.

 

Example:   Graph the linear equation y=3x

Solution:

 1. Find the x-intercept and the y-intercept.

To find an x-intercept: Let y=0 and solve for x.

y=3x

0=3x

{0}/3 = {3x}/3

0 = x

The x-intercept of this equation is (0,0)

To find a y-intercept: Let x=0 and solve for y.

y=3x

y=3(0)

y=0

The y-intercept of this equation is (0,0)

Since the x-intercept and the y-intercept are the same point and we need two distinct points to graph a line, we must find another ordered pair that is a solution to the equation.

Let x=1 and find the associated y value. (I chose x=1 but you could choose a different value)

y=3x

y=3(1)

y=3

Another ordered pair on the graph is (1,3)

2. Plot the x-intercept and the y-intercept.

3. Draw the line that connects the intercepts.

Example: Graphing a linear equation with intercepts.

Example: Graphing a linear equation with intercepts.

Example: Graphing a linear equation with intercepts.

Finding the Intercepts of a Circle Touching an Axis (Tangent to an axis)

An x-intercept is where the graph touches or crosses the x-axis.

A y-intercept is where the graph touches of crosses the y-axis.

To find an x-intercept: Let y=0 and solve for x.

To find an y-intercept: Let x=0 and solve for y.

 

Example:  Find the intercepts of the circle for the given equation.

(x-1)^2+(y+2)^2=4

Solution:

To find an x-intercept, let y=0 and solve for x.

(x-1)^2+(y+2)^2=4

(x-1)^2+(0+2)^2=4

(x-1)^2+(2)^2=4

(x-1)^2+4=4

(x-1)^2+4-4=4-4

(x-1)^2=0

sqrt{(x-1)^2}=sqrt{0}

x-1=0

x-1+1=0+1

x=1

This equation has one x-intercept. (1,0)

To find a y-intercept, let x=0 and solve for y.

(x-1)^2+(y+2)^2=4

(0-1)^2+(y+2)^2=4

(-1)^2+(y+2)^2=4

1+(y+2)^2=4

(y+2)^2+1-1=4-1

(y+2)^2=3

sqrt{(y+2)^2}=sqrt{3}

y+2=pm sqrt{3}

y+2=pm sqrt{3}

y+2-2=-2 pm sqrt{3}

y=-2 pm sqrt{3}

Approximately y=-3.732 and x=-0.2679

This equation has two y-intercepts. (0,-2 + sqrt{3}) and (0,-2 - sqrt{3})

A tangent line to a circle may be defined as a line that intersects the circle in a single point.

This circle is tangent to the x-axis since it is touching the x-axis in a single point.  The x-axis (y=0) is the tangent line for the point on the circle (1,0).

Example:  Find the intercepts of the circle for the given equation.

(x-3)^2+(y-1)^2=9

Solution:

To find an x-intercept, let y=0 and solve for x.

(x-3)^2+(y-1)^2=9

(x-3)^2+(0-1)^2=9

(x-3)^2+(-1)^2=9

(x-3)^2+1=9

(x-3)^2+1-1=9-1

(x-3)^2=8

sqrt{(x-3)^2}=sqrt{8}

x-3=sqrt{4*2}

x-3=pm 2sqrt{2}

x-3+3=3 pm 2sqrt{2}

x=3 pm 2sqrt{2}

Approximately x=0.1716 and x=5.828

This equation has two x-intercepts. (3+2sqrt{2},0) and (3-2sqrt{2},0)

To find a y-intercept, let x=0 and solve for y.

(x-3)^2+(y-1)^2=9

(0-3)^2+(y-1)^2=9

(-3)^2+(y-1)^2=9

9+(y-1)^2=9

(y-1)^2+9-9=9-0

(y-1)^2=0

sqrt{(y-1)^2}=sqrt{0}

y-1=0

y-1+1=0+1

y=1

This equation has one y-intercept. (0,1).

This circle is tangent to the y-axis since it is touching the y-axis in a single point.  The y-axis (x=0) is the tangent line for the point on the circle (0,1).