Completed square form quadratic graphs

[Probability]

I have;

f(x) = 3(x + 1)^2 - 12

If I draw a graph I am moving the + 1 to - 1 and downwards - 12 units, which will give me the new position of the quadratic graph.

If I wish to calculate the y intercept and the roots am I working the problem out in the correct way by;

3(x + 1)(x + 1) = x^2 + 2x + 2 = 3x^2 + 6x + 6 - 12 = 3x^2 + 6x - 6

If the above is correct, should I then divide through by 3;

x^2 + 3x - 3

If so

Factors

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

so f(0) = - 9

I am reasonably confident something is wrong as I would end up with no roots?

Edited

3x^2 + 6x - 9

follows that roots are;

x = 1 or x = 3

 

[JeffM]

I have;

f(x) = 3(x + 1)^2 - 12

If I draw a graph I am moving the + 1 to - 1 and downwards - 12 units, which will give me the new position of the quadratic graph.

If I wish to calculate the y intercept and the roots am I working the problem out in the correct way by;

3(x + 1)(x + 1) = x^2 + 2x + 2 = 3x^2 + 6x + 6 - 12 = 3x^2 + 6x - 6

If the above is correct, should I then divide through by 3;

x^2 + 3x - 3

If so

Factors

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

so f(0) = - 9

I am reasonably confident something is wrong as I would end up with no roots?

Edited

3x^2 + 6x - 9

follows that roots are;

x = 1 or x = 3

Are you required to solve the problem as a translation and expansion of the graph f(x) = x^2? If so, please give the exact wording of the exercise.

If not, this strikes me as an extremely convoluted way to deal with this problem. Restate the existing expression for f(x) into standard quadratic form and find the zeroes the regular way.

Do you know standard quadratic form?

Do you know how to find the zeroes of a quadratic in standard form?

 

[mmm4444bot]

f(x) = 3(x + 1)^2 - 12

I wish to calculate the y intercept

Substitute x=0, and do the arithmetic. In general, this is how we calculate y-intercepts.



Finding the roots is a separate process. To find the roots, set y=0 and solve for x

0 = 3(x + 1)^2 - 12

Add 12 to both sides, divide both sides by 3, take the square root of both sides, solve for x.

 

[HallsofIvy]

I have;

f(x) = 3(x + 1)^2 - 12

If I draw a graph I am moving the + 1 to - 1 and downwards - 12 units,

"moving +1 to -1" seems odd wording! I assume you mean that the vertex s at (-1, -12) which is correct.

which will give me the new position of the quadratic graph.

If I wish to calculate the y intercept and the roots am I working the problem out in the correct way by;

3(x + 1)(x + 1) = x^2 + 2x + 2 = 3x^2 + 6x + 6 - 12 = 3x^2 + 6x - 6

I have no idea what you are doing here. No, 3(x+1)(x+1) is NOT equal to \(\displaystyle x^2+ 2x+ 2\). \(\displaystyle (x+1)^2\) only would be equal to \(\displaystyle x^2+ 2x+ 1\), not "+2". Then multiplying that by 3 gives \(\displaystyle 3x^2+ 6x+ 3\). And subtracting 12 from that gives \(\displaystyle 3x^2+ 6x- 9\).
Part of your problem is that you are doing part of the calculation- which is perfectly reasonable- and then writing "=" between things that are NOT equal.

But why go to the work of multiplying when "completing the square" is a common way of solving quadratic equations- and you already have it in that form? The y-intercept, of course, is where x= 0 so \(\displaystyle y= 3(0+1)^2- 12= 3- 12= -9\). And the x-intercepts are where \(\displaystyle y= 3(x+1)^2- 12= 0\). That is, \(\displaystyle 3(x+1)^2= 12\) so \(\displaystyle (x+ 1)^2= 4\). That's easy to solve isn't it?

If the above is correct, should I then divide through by 3;

x^2 + 3x - 3

No, that's not at all right. 6 divided by 3 is 2, not 3. You should have \(\displaystyle x^2+ 2x- 3\)

If so

Factors

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

Where did this come from? You were dealing with \(\displaystyle x^2+ 3x- 3\) (though it should have been \(\displaystyle x^2+ 2x- 3\)), why now suddenly \(\displaystyle x^2- 9\)?

so f(0) = - 9

Yes, that's right, as I said before. That is what you get if you put x= 0 in \(\displaystyle 3(x^2- 3x- 3)\), 3(-3)= -9. But I have no idea what \(\displaystyle x^2- 9\) has to do with this!

I am reasonably confident something is wrong as I would end up with no roots?

Why do you say "no roots"? No roots to what equation?

Edited

3x^2 + 6x - 9

follows that roots are;

x = 1 or x = 3

No, it doesn't. \(\displaystyle 3(1)^2+ 6(1)- 9= 0\) but \(\displaystyle 3(3)^2+ 3(3)- 9= 36- 9= 27\), not 0. As I said before, multiplying out and then factoring is the hard way. Directly from the "complete square" form you get \(\displaystyle (x+1)^2= 4\) so that \(\displaystyle x+ 1=\pm 2\). Taking +2, we have x+1= 2 so x= 1 and taking -2, we have x+1= -2 so x= -3.

 

[Probability]

"moving +1 to -1" seems odd wording! I assume you mean that the vertex s at (-1, -12) which is correct.


I have no idea what you are doing here. No, 3(x+1)(x+1) is NOT equal to \(\displaystyle x^2+ 2x+ 2\). \(\displaystyle (x+1)^2\) only would be equal to \(\displaystyle x^2+ 2x+ 1\), not "+2". Then multiplying that by 3 gives \(\displaystyle 3x^2+ 6x+ 3\). And subtracting 12 from that gives \(\displaystyle 3x^2+ 6x- 9\).
Part of your problem is that you are doing part of the calculation- which is perfectly reasonable- and then writing "=" between things that are NOT equal.

But why go to the work of multiplying when "completing the square" is a common way of solving quadratic equations- and you already have it in that form? The y-intercept, of course, is where x= 0 so \(\displaystyle y= 3(0+1)^2- 12= 3- 12= -9\). And the x-intercepts are where \(\displaystyle y= 3(x+1)^2- 12= 0\). That is, \(\displaystyle 3(x+1)^2= 12\) so \(\displaystyle (x+ 1)^2= 4\). That's easy to solve isn't it?


No, that's not at all right. 6 divided by 3 is 2, not 3. You should have \(\displaystyle x^2+ 2x- 3\)


Where did this come from? You were dealing with \(\displaystyle x^2+ 3x- 3\) (though it should have been \(\displaystyle x^2+ 2x- 3\)), why now suddenly \(\displaystyle x^2- 9\)?


Yes, that's right, as I said before. That is what you get if you put x= 0 in \(\displaystyle 3(x^2- 3x- 3)\), 3(-3)= -9. But I have no idea what \(\displaystyle x^2- 9\) has to do with this!


Why do you say "no roots"? No roots to what equation?


No, it doesn't. \(\displaystyle 3(1)^2+ 6(1)- 9= 0\) but \(\displaystyle 3(3)^2+ 3(3)- 9= 36- 9= 27\), not 0. As I said before, multiplying out and then factoring is the hard way. Directly from the "complete square" form you get \(\displaystyle (x+1)^2= 4\) so that \(\displaystyle x+ 1=\pm 2\). Taking +2, we have x+1= 2 so x= 1 and taking -2, we have x+1= -2 so x= -3.

Thanks for the efforts put in to explain this to me. I must admit that this branch of maths is brand new to me and have little to no experince in it. For the course I am on and this level of pitch as an introduction to these functions I think is a bit OTT. When I refer back to my last maths course material there is no introduction to this subject, and the style being used is not readily found in any maths books I have, which is making the subject overwhelming to get a grip on from first principles.

I appreciate the work you have completed above but the steps are a little hard to follow.

 

[JeffM]

Thanks for the efforts put in to explain this to me. I must admit that this branch of maths is brand new to me and have little to no experince in it. For the course I am on and this level of pitch as an introduction to these functions I think is a bit OTT. When I refer back to my last maths course material there is no introduction to this subject, and the style being used is not readily found in any maths books I have, which is making the subject overwhelming to get a grip on from first principles.

You did not answer my question. What does the problem say exactly? It is very easy to solve non-graphically.

\(\displaystyle Find\ the\ x\ and\ y\ intercepts\ of\ f(x) = 3(x + 1)^2 - 12.\) Is this the problem? The x intercepts may be called zeroes or roots.

To find the y intercept of ANY FUNCTION, you simply evaluate it at x = 0. So a function does not have a y intercept if the function is not defined at x = 0. If the function is defined at x = 0, then the function has a unique y intercept.

\(\displaystyle f(0) = 3(0 + 1)^2 - 12 = 3(1)^2 - 12 = 3 * 1 - 12 = 3 - 12 = - 9.\)

To find x intercepts of ANY FUNCTION, you solve the equation f(x) = 0. So a function may have no x intercepts, exactly one x intercept, or more than one x intercept.

\(\displaystyle f(x) = 0 \implies 3(x + 1)^2 - 12 = 0 \implies 3(x^2 + 2x + 1) - 12 = 0 \implies 3x^2 + 6x + 3 - 12 = 0 \implies 3x^2 + 6x - 9 = 0.\)

You have an equation in standard quadratic form. Do you know how to solve it?

 

[Probability]

You did not answer my question. What does the problem say exactly? It is very easy to solve non-graphically.

\(\displaystyle Find\ the\ x\ and\ y\ intercepts\ of\ f(x) = 3(x + 1)^2 - 12.\) Is this the problem? The x intercepts may be called zeroes or roots.

To find the y intercept of ANY FUNCTION, you simply evaluate it at x = 0. So a function does not have a y intercept if the function is not defined at x = 0. If the function is defined at x = 0, then the function has a unique y intercept.

\(\displaystyle f(0) = 3(0 + 1)^2 - 12 = 3(1)^2 - 12 = 3 * 1 - 12 = 3 - 12 = - 9.\)

To find x intercepts of ANY FUNCTION, you solve the equation f(x) = 0. So a function may have no x intercepts, exactly one x intercept, or more than one x intercept.

\(\displaystyle f(x) = 0 \implies 3(x + 1)^2 - 12 = 0 \implies 3(x^2 + 2x + 1) - 12 = 0 \implies 3x^2 + 6x + 3 - 12 = 0 \implies 3x^2 + 6x - 9 = 0.\)

You have an equation in standard quadratic form. Do you know how to solve it?

Thanks Jeff, in my first thread I was struggling to work it out but managed to find that I got to 3x^2 + 6x - 9, and you have confirmed this is correct and my roots are as I said originally x = 1 and x = 3.

Thanks for the effort and back up.

The original question which I still don't profess to understand is asking;

Explain how the parabola that is the graph of f can be obtained from the graph of y = x^2 by using appropriate translations and scaling.

I am lost what the question is asking for by saying explain I am confident however that it does not want me to right words about it?

 

[phoenix9124]

Thanks Jeff, in my first thread I was struggling to work it out but managed to find that I got to 3x^2 + 6x - 9, and you have confirmed this is correct and my roots are as I said originally x = 1 and x = 3.

Thanks for the effort and back up.

The original question which I still don't profess to understand is asking;

Explain how the parabola that is the graph of f can be obtained from the graph of y = x^2 by using appropriate translations and scaling.

I am lost what the question is asking for by saying explain I am confident however that it does not want me to right words about it?

I guess they mean that you can obtain g(x) = 3(x+1)^2 - 12
by taking certain steps starting from f(x) = x^2

Taking the standard form: h(x) = a(x+b)^2 - c
With b and c being the translations (b in -x, c in +y) and a the scaling factor.
Shifting the graph 1 step (in x direction) to the left, would mean b should increase with 1.

It's very helpfull to use plotters on the internet to just play with these numbers and see what happens when you change a,b,c in h(x).

As i've done here: http://fooplot.com/#W3sidHlwZSI6MCw...1IiwiLTE1IiwiNiJdLCJncmlkIjpbIjEiLCIxIl19XQ--

Jeez... long URL.

Good luck.

 

[JeffM]

Thanks Jeff, in my first thread I was struggling to work it out but managed to find that I got to 3x^2 + 6x - 9, and you have confirmed this is correct and my roots are as I said originally x = 1 and x = 3. Actually, this is wrong. The roots or zeroes or x-intercepts are x = MINUS 3 and x = PLUS 1. Do you understand why? Try substituting your proposed solutions into the original equation.

Thanks for the effort and back up.

The original question which I still don't profess to understand is asking;

Explain how the parabola that is the graph of f can be obtained from the graph of y = x^2 by using appropriate translations and scaling.

I am lost what the question is asking for by saying explain I am confident however that it does not want me to right words about it?

Aha, the light dawns for me now that I understand what the question is.

Consider the two functions \(\displaystyle g(x) = 3 * f(x).\) For every value of x, g(x) is three times f(x). 3 is the relevant scaling fator. The scale at which we are looking at things has been tripled. Make sense?

Now let's think about \(\displaystyle m(x) = f(x + 1)\) The graphs of m(x) and f(x) will look exactly the same except the graph of m(x) will be shifted to the left by 1. That is:

\(\displaystyle f(x) = m(x + 1)\ and f(x) = x^2 \implies f(-3) = (-3)^2 = 9 = (-3)^2 = (-4 + 1)^2 = m(- 4).\ And\ f(2) = 2^2 = 4 = (2)^2 = (1 + 1)^2 = m(1).\)

"Translate" comes from the Latin words "trans" = "across or along" and "latus" = "be carried," and so "translate" originally meant "to be carried across or along." When we translate from one language to another, the meaning is "carried across" the barrier of language. In the example above, f(x) is carried "along" the x-axis to the left by 1 to make m(x). Make sense?

\(\displaystyle f(x) = p(x) + a,\ where\ a\ is\ negative,\ zero,\ or\ positive.\) This is a translation "along" the y-axis. The graphs will look exactly the same except their position relative to the x-axis will differ. Make sense? So

\(\displaystyle f(x) = p(x) - 12\ and\ f(x) = x^2 \implies f(0) = 0\ and\ p(0) = - 12.\)

To get p(x), f(x) has been carried along the y-axis downward by 12.

So scaling and translations allow you to transform functions or graphs into related but different functions or graphs.The whole purpose is getting you to think generally about a family of related functions, whose graphs have the same general shape, but are positioned differently, rather than a single function. (To drive you nuts, you will later discover another standard transformation called a rotation and even weirder transformations called homeomorphisms and diffeomorphisms, which I am far too ignorant to discuss intelligently. The whole idea about all classes of transformations is that something important is common to any particular class of transformation.)

If this still does not make sense to you, I must admit that I am out of explicatory ideas, and I shall let someone else try to explain it to you. Sorry.

 

[Probability]

Aha, the light dawns for me now that I understand what the question is.

Consider the two functions \(\displaystyle g(x) = 3 * f(x).\) For every value of x, g(x) is three times f(x). 3 is the relevant scaling fator. The scale at which we are looking at things has been tripled. Make sense?

Now let's think about \(\displaystyle m(x) = f(x + 1)\) The graphs of m(x) and f(x) will look exactly the same except the graph of m(x) will be shifted to the left by 1. That is:

\(\displaystyle f(x) = m(x + 1)\ and f(x) = x^2 \implies f(-3) = (-3)^2 = 9 = (-3)^2 = (-4 + 1)^2 = m(- 4).\ And\ f(2) = 2^2 = 4 = (2)^2 = (1 + 1)^2 = m(1).\)

"Translate" comes from the Latin words "trans" = "across or along" and "latus" = "be carried," and so "translate" originally meant "to be carried across or along." When we translate from one language to another, the meaning is "carried across" the barrier of language. In the example above, f(x) is carried "along" the x-axis to the left by 1 to make m(x). Make sense?

\(\displaystyle f(x) = p(x) + a,\ where\ a\ is\ negative,\ zero,\ or\ positive.\) This is a translation "along" the y-axis. The graphs will look exactly the same except their position relative to the y-axis will differ. Make sense? So

\(\displaystyle f(x) = p(x) - 12\ and\ f(x) = x^2 \implies f(0) = 0\ and\ p(0) = - 12.\)

To get p(x), f(x) has been carried along the y-axis downward by 12.

So scaling and translations allow you to transform functions or graphs into related but different functions or graphs.The whole purpose is getting you to think generally about a family of related functions, whose graphs have the same general shape, but are positioned differently, rather than a single function. (To drive you nuts, you will later discover another standard transformation called a rotation and even weirder transformations called homeomorphisms and diffeomorphisms, which I am far too ignorant to discuss intelligently. The whole idea about all classes of transformations is that something important is common to any particular class of transformation.)

If this still does not make sense to you, I must admit that I am out of explicatory ideas, and I shall let someone else try to explain it to you. Sorry.

Thanks Jeff I think I am understanding it now, and yes the root is as you say - 3 and not 3 as I thought.

 

[Probability]

I guess they mean that you can obtain g(x) = 3(x+1)^2 - 12
by taking certain steps starting from f(x) = x^2

Taking the standard form: h(x) = a(x+b)^2 - c
With b and c being the translations (b in -x, c in +y) and a the scaling factor.
Shifting the graph 1 step (in x direction) to the left, would mean b should increase with 1.

It's very helpfull to use plotters on the internet to just play with these numbers and see what happens when you change a,b,c in h(x).

As i've done here: http://fooplot.com/#W3sidHlwZSI6MCw...1IiwiLTE1IiwiNiJdLCJncmlkIjpbIjEiLCIxIl19XQ--

Jeez... long URL.

Good luck.

Thanks
phoenix the link is a great help.