Quadratic Equation: Analyzing Graph of y=(5-2x)^2+1

AI Thread Summary
The discussion focuses on analyzing the transformations of the quadratic equation y=(5-2x)^2+1 compared to the standard form y=x^2. Key transformations include a vertical translation of 1 unit upward and a horizontal translation of 2.5 units to the right. The factor of 2 in the term 2x indicates a horizontal compression, which results in a dilation factor of 4 parallel to the y-axis. Confusion arises regarding the dilation factor, with one participant suggesting a method to rewrite the function for clarity. The proposed method involves expanding the squared term and rewriting the function in a standard transformation form, which could be applicable to similar problems.
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The graph of y=(5-2x)^2+1. What are the transformations that have occurred from y=x^2.

Im really confused with the transformations here.
 
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mainly confused about the dilation factor.
 
For y=f(x)+b, the effect of changes in b is to translate the graph of y=f(x) vertically through b units, so now you know what the +1 at the end does.

For y=f(kx), k>0, the effect of changes in k is to horizontally stretch or compress the graph of y=f(x) by a factor of 1/k , so that takes care of the 2x.

As you can see, everything in your transformed graph, including the minus sign does one simple translation, and combined you get your graph.

They're nicely summarized here and here's a more comprehensive explanation ;)

The dilation is due to the 2x, since every y-value is now doubled, it rises faster and so appears to have shrunk.
 
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The answer to this question is: dilation factor of 4 parallel to the y axis, translation of 2.5 to the right and 1 unit upwards.
Okay so what I did was change it into this y=[-2(x-5/2)]^2+1, but that gives me a dilation factor of 2 parallel to the y axis. What am I doing wrong?
 
This is actually really simple:
1. expand the squared quantity
2. Try to rewrite this function in the form of y=af(bx+c)+d with f(x)=x^2
3. Use your textbook to interpret this answer.
 
thankyou thankyou ! would this method work with all these type of questions?
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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