Combining Transformations; Completing the Square

In summary, the function y = 2x2 + x +1 is transformed by being stretched vertically by a factor of 2, stretched horizontally by a factor of 1/3, and translated 2 units to the right and 4 units down. The equation of the transformed function is y= 2(2(3(x-1/4)2)-15/16).
  • #1
trulyfalse
35
0
Hello PF!

Homework Statement


The graph of the function y = 2x2 + x +1 is stretched vertically about the x-axis by a factor of 2, stretched horizontally about the y-axis by a factor of 1/3 and translated 2 units right and 4 units down. Write the equation of the transformed function

Homework Equations


y= af(1/b(x-h))+k

The Attempt at a Solution


First I completed the square of y = 2x2 + x +1:
2x2 + x +1 = y
2(x2+1/2x+1/16-1/16)+1=y
2(x+1/4)2+14/16=y

Then, using mapping notation I calculated what the new x and y coordinates would be on the transformed function:
(x,y) → (1/3x+2,2y-4)
Therefore, the point (-2,7) → (4/3,10)

Using the aforementioned equation I transformed the function:
a=2
b=1/3
h=2
k=-4

y=2(2(3(x-7/4)2)-25/8)

However, when I input the new function into my calculator I received the values (-2,7) → (4/3,53/9) which is demonstrated to be incorrect. Is there a way to solve this problem without putting the function in the form y= af(1/b(x-h))+k? If possible, can anyone show me how to derive the correct equation for the transformed function after completing the square of the function? Thanks!
 
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  • #2
trulyfalse said:
Hello PF!

Homework Statement


The graph of the function y = 2x2 + x +1 is stretched vertically about the x-axis by a factor of 2, stretched horizontally about the y-axis by a factor of 1/3 and translated 2 units right and 4 units down. Write the equation of the transformed function

Homework Equations


y= af(1/b(x-h))+k


The Attempt at a Solution


First I completed the square of y = 2x2 + x +1:
2x2 + x +1 = y
2(x2+1/2x+1/16-1/16)+1=y
2(x+1/4)2+14/16=y
Almost. 1- 1/16= 15/16, not 14/16.

Thern, using mapping notation I calculated what the new x and y coordinates would be on the transformed function:
(x,y) → (1/3x+2,2y-4)
Therefore, the point (-2,7) → (4/3,10)
where did the "1/3" come from? If you are comparing y= 2(x+ 1/4)2+ 15/16 to y= x2 then x changes to x+ 1/4 and y changes to 2y+ 15/16.

Using the aforementioned equation I transformed the function:
a=2
b=1/3
h=2
k=-4

y=2(2(3(x-7/4)2)-25/8)

However, when I input the new function into my calculator I received the values (-2,7) → (4/3,53/9) which is demonstrated to be incorrect. Is there a way to solve this problem without putting the function in the form y= af(1/b(x-h))+k? If possible, can anyone show me how to derive the correct equation for the transformed function after completing the square of the function? Thanks!
 

1. What is the purpose of combining transformations?

Combining transformations allows us to simplify complex functions and make them more manageable. This is especially useful in solving equations or graphing functions.

2. How do you combine transformations?

To combine transformations, we perform each transformation in order starting from the innermost function. This means that we first apply the inside function and then the outside function.

3. What are the steps for completing the square?

The steps for completing the square are as follows:
1. Move the constant term to the other side of the equation
2. Divide the coefficient of the x-term by 2 and square the result
3. Add the squared result to both sides of the equation
4. Factor the trinomial on the left side of the equation
5. Take the square root of both sides of the equation
6. Solve for x

4. Why do we use completing the square?

We use completing the square to solve quadratic equations that cannot be easily factored or solved using other methods. This technique allows us to convert a quadratic expression into a perfect square trinomial, making it easier to solve.

5. How can we use completing the square to graph a quadratic function?

Completing the square can help us graph a quadratic function by finding the vertex of the parabola, which is the highest or lowest point on the graph. The x-coordinate of the vertex can be found by using the formula -b/2a, where a and b are the coefficients of the quadratic function in standard form. The y-coordinate can be found by substituting the x-value into the original function.

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