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Drawing log graphs

November 5, 2010 Leave a comment

First lets consider the basic log graph y=log_{10} x

Log graph features

The important features

  • The graph has an asymptote of x=0
  • It intersects the x axis at x=1, because the log of 1 is zero
  • It passes through the point (10,1), because 10^1=1.  Notice 10 is the base
  • The curve is always increasing, but at a slower and slower rate

Conclusions

The asymptote will always be x=0 unless the graph is translated \Rightarrow y=log_{10}(x+a) has an asymptote at x=-a

y=1 when x equals the base number unless the graph has been translated \Rightarrow y=log_a(x) goes through (a,1)

The graph goes through (1,0) unless translated \Rightarrow y=log_{10}(x+a)+b goes through the point (10-a,1+b)

Examples

For each of the function below give the asymptote, the coordinates where the point (1,0) has moved to and the coordinates of where the point (base,1) has moved to.

y=log_5x

The asymptote has not changed since we have not added to x so the asymptote is x=0

The point related to the base is (5,1)

The point (1,0) has not moved since we have not added to x or y

Confirms claims above

y=log_{10}x+2

The asymptote has not changed since we have not added to x so the asymptote is x=0.

The point related to the base is (10,3) because we have added 2 to the function so the graph have moved up 2.

The point (1,0) has moved to (1,2) since we have added 2 to the function so the graph have moved up 2.

Confirms points made above

y=3log_{10}(x-1)+2

The asymptote is x=1 because we have subtracted 1 from x and this shifts the graph 1 place to the right.

The point related to the base is (11,5) because the y values are 3 times bigger and then we add 2.

The point (1,0) has moved to (1,2) since we need to multiply y by 3 (no effect here) and then add 2.

Confirms points made above

Questions

Categories: Graphing

Finding the mid point between to coordinates

November 5, 2010 Leave a comment

You can see that the horizontal position of the mid point is half way between the horizontal positions of A and B.  Also you can see that the vertical position of the mid point is half way between the vertical positions of A and B.

\therefore mid point = (X_{average},Y_{average})

Example

Find the mid point of (3,8) and (7,-3)

Mid point =\left(\frac{3+7}{2},\frac{8+-3}{2}\right)=\left(5,\frac{5}{2}\right)

Categories: Graphing

Gradient between two points

November 5, 2010 Leave a comment

The gradient between two points is the gradient of the line that passes through them.

Two points

Gradient = \frac{rise}{run}

The rise is the change is y = y_2-y_1

The run is the change is x = x_2-x_1

So the gradient = \frac{y_2-y_1}{x_2-x_1}

It does not matter if you do the first coordinate minus the second or the other way round provided that you are consistent.

Examples

Find the gradient between (3, 6) and (7,14)

gradient = \frac{14-6}{7-3}=\frac{8}{4}=2


Find the gradient between (2,7) and (5,6)

gradient = \frac{6-7}{5-2}=-\frac{1}{3}


Show that the quadrilateral with vertices A(0,0), B(6,4), C(4,6) and D(1,4) is a trapezium.

To be a trapezium the shape must have a pair of parallel lines, which means a pair of lines with the same gradient.

Gradient AB=\frac{4-0}{6-0}=\frac{2}{3}

Gradient BC=\frac{6-4}{4-6}=-1

Gradient CD=\frac{6-4}{4-1}=\frac{2}{3}

Gradient DA=\frac{4-0}{1-0}=4

Since AB and CD have the same gradient the shape must be a trapezium

Categories: Graphing

Distance between two points

November 2, 2010 Leave a comment

Let’s start by considering the points (x_1,y_1) and (x_2,y_2)

Coordinate Pythagoras

We have used the two points to construct a right-angled triangle.

The horizontal distance is equal to x_2-x_1

The vertical distance is equal to y_2-y_1

So using Pythagoras’ theorem we can say the distance between the points =\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

If we define the change in x as \delta x and the change in y as \delta y the the distance become s=\sqrt{(\delta x)^2+(\delta y)^2}

Examples

Find the distance between the two points (3,7) and (7,-2)

\delta x = 3-7=-4.  Since this is the length of a side of a triangle we will positive 4.

\delta y = 7--2=9

\therefore distance = \sqrt{4^2+7^2}=\sqrt{65}


A triangle is made with the three points A=(2,8),B=(5,12) and C=(6,11).  Show that the triangle is an isosceles triangle.

 

An isosceles triangle has two sides of equal length, if we can show that two of our sides are equal then we have proved it is isosceles.

The length AB =\sqrt{3^2+4^2}=5

The length BC =\sqrt{1^2+1^2}=\sqrt{2}

The length AC=\sqrt{4^2+3^2}=5

Since the length AB = the length AC, the triangle is isosceles.

Categories: Graphing

Lines intersecting circles

October 29, 2010 Leave a comment

When a line intersects a circle it will either have one point of contact in which case it is a tangent to the circle or it will cut it in two places.

To find the points of contact between a line and a circle

  1. If needed, rearange the equation of the line to make x or y the subject
  2. Substitute into the circle equation and solve the resulting quadratic
  3. Substitute the solutions into the equation of the line to get the required coordinates

Example

Find where the line y=2x+1 intersects with the circle (x+2)^2+(y-1)^2=5

When we substitute y into the circle we get (x+2)^2 + ((2x+1)-1)^2=5

\therefore (x+2)^2+(2x)^2=5 \Rightarrow x^2+4x+4+4x^2=5 \Rightarrow 5x^2+4x-1=0

\therefore (5x-1)(x+1)=0 \Rightarrow x=\frac{1}{5} or x = -1

If x=\frac{1}{5}, y=\frac{7}{5}

If x=-1, y=-1

\therefore the points of intersection are (\frac{1}{5}, \frac{7}{5}) and (-1,-1)

Categories: Graphing

Where two lines intersect

October 29, 2010 Leave a comment

Using the substitution method to find where two lines intersect

  1. Rearrange one of the equations to make either x or y the subject
  2. Substitute for either x or y depending on which one you made the subject
  3. Solve the resulting equation
  4. Substitute this value into one of the original equations to find the value of the other variable

Example

Find the coordinates where the lines y=3x+5 and 3x+2y=28 intersect

Since the first equation already has y as the subject we will substitute this into the second

\therefore 3x+2(3x+5)=28 \Rightarrow 9x+10=28 \Rightarrow x=2

We now substitute this back into the first (on this occassion it is the easiest option)

and get y=3 \times 2+5=11

\therefore the point of intersection in (2,11)

Categories: Graphing

Finding Lines using y-y1=m(x-x1)

October 27, 2010 Leave a comment

A useful way of finding the equation of a line is to use the formula y-y_1=m(x-x_1), where m is the gradient of the line and x_1 and y_1 are taken from the given point (x_1,y_y)

Finding the equation of the line given the gradient and a point

  1. Get the values of m, x_1, y_1
  2. Substitute into the formula
  3. Rearrange the formula to the desired form

Example

Find the line with a gradient of -2 that passes through the point (2,6)

So m=-2, x_1=2, y_1=6

\therefore y-6=-2(x-2) \Rightarrow y=-2x+10

Finding the equation of the line given a parallel line and a point

  1. Use the gradient of the parallel line for m
  2. Get the values of x_1 and y_1 from the given point
  3. Substitute into the formula
  4. Rearrange the formula to the desired form

Example

Find the equation of the line which  is parallel to y=3x+1 and passes through the point (2,1)

So m=3, x_1=2, y_1=1

\therefore y-1=3(x-2) \Rightarrow y=3x-5

Finding the equation of the line given a perpendicular line and a point

  1. Get the gradient of the perpendicular line and use the negative reciprocal for m
  2. Get the values of x_1 and y_1 from the given point
  3. Substitute into the formula
  4. Rearrange the formula to the desired form

Example

Find the equation of the line which  is perpendicular to y=-4x+7 and passes through the point (8,1)

The gradient of the given line is -4, so the perpendicular gradient is \frac{1}{4} \therefore m=\frac{1}{4}

From the point x_1 = 8 and y_1 = 1

\therefore y-1=\frac{1}{4}(x-8) \Rightarrow y=\frac{1}{4}x-1

Finding the equation of the line given two points

  1. Find the gradient from the points using \frac{rise}{run} or \frac{\delta y}{\delta x}
  2. Use either point in the formula, your choice
  3. Rearrange the formula to the desired form

Example

Find the line which passes through the points (-4, 7) and (3,10)

m = gradient = \frac{10-7}{3--4}=\frac{3}{7}

I will use the point (3,10), because both values are positive, but remember you would get the same result if you used (-4, 7)

\therefore x_1 = 3 and y_1 = 10

So substituting gives y-10=\frac{3}{7}(x-3) \Rightarrow y=\frac{3}{7}x+\frac{61}{7}

Finding the equation of the Perpendicular bisector between two points

  1. Find the gradient between the points and use the negative reciprocal for m
  2. Find the mid-point between the two points and use this for x_1 and y_1
  3. Substitute into the formula
  4. Rearrange the formula to the desired form

Example

Find the perpendicular bisector between the points (4, 6) and (2,10)

Gradient between the points is \frac{10-6}{2-4}=-2

\therefore m =\frac{1}{2}

The mid-point if \left(\frac{4+2}{2},\frac{6+10}{2}\right)=(3,8)

\therefore x_1 = 3 and y_1 = 8

So substituting gives y-8=\frac{1}{2}(x-3) \Rightarrow y=\frac{1}{2}x+\frac{13}{2}

Categories: Graphing