Explain in detail where the formula for the difference quotient comes
First of all, the way this works is that we don't have any numbers so we have to come up with the formula replacing the numbers for letters. x will represent the values on the x axis. h will represent the change in x.
First we get a point on the graph. that will be our starting point. That point will be (x, f(x)) it's this because the graph starts at a certain x point and the y point is related to the x point chosen therefore y is related to the function of x. now we have to find the second point. in the moment you move from x to the right, its not x anymore. It's x plus the change on x. therefore it'd be x+h, while the height would still be in relation to the its x value which in this case it's x+h that leaving us with the point (x+h, f(x+h)).
Now we have to find the slope between these two points. The formula to find the slope will still be the same m=(y2-y1)/(x2-x1). when we plug in the values the formula would be
[f(x+h) - f(x)]/[(x+h)-(x)] the top will stay the same while the in the bottom the x's will cancel leaving just h at the bottom. leaving us with the difference quotient formula.
works cited
http://images.tutorvista.com/cms/images/39/difference-quotient-formula.png
http://cis.stvincent.edu/carlsond/ma109/DifferenceQuotient_images/IMG0470.JPG
Friday, June 6, 2014
Monday, June 2, 2014
I/D#2 - Unit O Concept 7-8: Deriving Special Right Triangles
How can we derive the 30-60-90 triangle from an equilateral triangle with a side length of 1?
First we have the triangle as shown above.
Then we divide the triangle in half making one of the 60 degree angles become two 30 degree angles. By doing this we also create two 90 degree angles. Now we have to find the missing side being the dotted line. we can do this with the pythagorean theorem. With the bottom being 1/2 and the side being 1. The answer should be y = rad3 / 2. Then we multiply all the values of the triangle by two so that we don't have any fractions.
This should be the final result of the triangle.
How can we derive the 45-45-90 triangle from an square with a side length of 1?
First you draw the square and label each one of the sides one.
Then you make a diagonal dividing the square into two triangles. You may notice that these are the two triangles are the triangles that you're looking for (45, 45, 90)
Then you use the pythagorean theorem to find the length of r. this resulting to be rad2.
Then when you get that you multiply it by n so that you get the derivation of the 45 - 45 - 90 triangle, as shown below
Something I never noticed before about special right triangles is…
That they are derived from another shape and the angles come from there.
Being able to derive these patterns myself aids in my learning because…
now I where the formulas and the angles come form. if one day I don't remember how the concept works I can always come back here and know the basics of the special right triangles.
First we have the triangle as shown above.
Then we divide the triangle in half making one of the 60 degree angles become two 30 degree angles. By doing this we also create two 90 degree angles. Now we have to find the missing side being the dotted line. we can do this with the pythagorean theorem. With the bottom being 1/2 and the side being 1. The answer should be y = rad3 / 2. Then we multiply all the values of the triangle by two so that we don't have any fractions.
How can we derive the 45-45-90 triangle from an square with a side length of 1?
First you draw the square and label each one of the sides one.
Then you make a diagonal dividing the square into two triangles. You may notice that these are the two triangles are the triangles that you're looking for (45, 45, 90)
Then you use the pythagorean theorem to find the length of r. this resulting to be rad2.
Then when you get that you multiply it by n so that you get the derivation of the 45 - 45 - 90 triangle, as shown below
Something I never noticed before about special right triangles is…
That they are derived from another shape and the angles come from there.
Being able to derive these patterns myself aids in my learning because…
now I where the formulas and the angles come form. if one day I don't remember how the concept works I can always come back here and know the basics of the special right triangles.
SP #7: Unit Q concept 7. Finding all trig functions.
Subscribe to:
Posts (Atom)