Quadratic Parabola equations

hmmm, this is slightly trickier as it includes a y which I am not familiar with.
I'm not a hundred percent sure so I will give two estimates.

#1
5x - 2 = y
5(x-2)=y
5x - 10 = y
5x = y + 10
y= 5x + 10


#2
5x - 2 = y
5x = y+2
x = y+2 /5


I have a feeling this might be wrong. To me it looks a lot better on the first guess. I can never be sure though.
Is the first one x=-2?? I just noticed that 5 x -2 = -10, if you use the other 10 that is positive, they cancel each other out.
To me it makes more sense that way

The first is obviously wrong; you didn't even end up with "x = ..." at all, and what you do say is incompatible with what you started with.

The second is correct if you add in parentheses that are needed to make it say what you intend: x = (y+2)/5 .

One way to check your answer is to plug it into the original equation: 5[(y+2)/5] - 2 = y. If that simplifies down to something that is always true (y = y), then you're good.

That is a little awkward, so another thing you can do is what I call a "spot check": pick a couple values of x and see whether the y you get satisfies the other equation. If I take x = 3, then y = 5(3) - 2 = 13; putting that into your solution, x = (13 + 2)/5 = 3, which is good.

If you tried this with your first answer, you'd quickly see how wrong it is.
 
JEFFM you are really confusing. I know nothing about maths but you continue like I should understand it. I never learnt algebra but I learnt implicit differentiation because that is part of my study. It's like learning a new language but instead your not bred into it. That doesn't mean it is impossible to learn. A child can pick things up a lot quickly but it takes him years to understand it. An adult however only needs a few practices and has to maintain it otherwise it is lost. I may not have the full understanding yet because I am just getting into the flow of learning how the process goes. At the same time Dr.Peterson makes a point that I tend to be rushing things without thinking it through and letting all the information sink in (because honestly I am rushing as I am trying to get this study completed as quick as possible before February). I notice you are trying to stress the point that I can't do this. I don't know where your discouragement comes from but that doesn't look like it can help anyone. However I give thanks to the points you did make earlier:

\(\displaystyle 18x^2\) means that you multiply the number x by itself and then multiply that product by 18.

\(\displaystyle 18^2\) means that you multiply 18 by itself.

I didn't know that but now I understand thanks for sharing.


The first is obviously wrong; you didn't even end up with "x = ..." at all, and what you do say is incompatible with what you started with.

The second is correct if you add in parentheses that are needed to make it say what you intend: x = (y+2)/5 .

One way to check your answer is to plug it into the original equation: 5[(y+2)/5] - 2 = y. If that simplifies down to something that is always true (y = y), then you're good.

That is a little awkward, so another thing you can do is what I call a "spot check": pick a couple values of x and see whether the y you get satisfies the other equation. If I take x = 3, then y = 5(3) - 2 = 13; putting that into your solution, x = (13 + 2)/5 = 3, which is good.

If you tried this with your first answer, you'd quickly see how wrong it is.

Thanks Dr.Peterson, ok I didn't understand all that you said but I think I get the idea.
For the equation 5x -2 =y

To solve equation for y:
y = 2
because if you solve for y, the x must be 0 so..
5(0) -2 = y
5 times 0 is 0 right? so what is left: -2 = y.
swap them around and you get...
y = 2

To solve the equation for x:
x = 0.4
because if you solve for x, the y must then be cancelled out so..
5x - 2 = 0
5x = 2
x= 2/5
x = 0.4

To check if this is correct:
5 x 0.4 = 2 - 2= 0

Hopefully that is right?
If that is the case, you asked earlier about finding the y for x^2 = -28 (y - 8.5)
If the y is being solved then the x must be 0 so...
0^2 = -28 (y - 8.5)
y= -28 - 8.5
y= 28 + 8.5
y = 36.5

to check if this equals to 0
x^2 = -28 (36.5 - 8.5)
36.5 - 8.5 = 28
-28 + 28
28 - 28 = 0

I really do hope thats correct. If not, then I might need more practice.
Thanks again for your time and help Dr.Peterson! :)
 
For the equation 5x -2 =y

To solve equation for y:
y = 2
because if you solve for y, the x must be 0 so..
5(0) -2 = y
5 times 0 is 0 right? so what is left: -2 = y.
swap them around and you get...
y = 2

No, you can't just assume that x=0. Solving for y means finding an expression in terms of x for the value of y.

This equation is already solved for y! There is no work to do.

To solve the equation for x:
x = 0.4
because if you solve for x, the y must then be cancelled out so..
5x - 2 = 0
5x = 2
x= 2/5
x = 0.4

To check if this is correct:
5 x 0.4 = 2 - 2= 0

Hopefully that is right?

Solving for x is what you did before, correctly (in your second attempt), as we've told you. Again, you can't assume y is 0.

If that is the case, you asked earlier about finding the y for x^2 = -28 (y - 8.5)
If the y is being solved then the x must be 0 so...
0^2 = -28 (y - 8.5)
y= -28 - 8.5
y= 28 + 8.5
y = 36.5

to check if this equals to 0
x^2 = -28 (36.5 - 8.5)
36.5 - 8.5 = 28
-28 + 28
28 - 28 = 0

I really do hope thats correct. If not, then I might need more practice.
Thanks again for your time and help Dr.Peterson! :)

Again, you can't assume that y is zero; and you made another silly mistake in your first step anyway, and another in your check.

Try again, rearranging the equation in much the same way you solved 5x - 2 = y for x before.
 
Ok so I can't assume y is 0.
If y isn't 0 then it has to be a number.

5x - 2 = y

5x = y + 2
x = y+2 /5
2 x 5 = 10/5
x = 2

so...

5 x 2 = 10 - 2 = 8
y = 8??
 
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Ok so I can't assume y is 0.
If y isn't 0 then it has to be a number.

5x - 2 = y

5x = y + 2
x = y+2 /5
2 x 5 = 10/5
x = 2

so...

5 x 2 = 10 - 2 = 8
y = 8??

The answer is what you said here, and had said before (written correctly): x = (y + 2)/5.

As I said, solving for x means expressing x as an expression in terms of y, which is what this is.

With this solved formula, for ANY value of y, you can determine what x is. If y=0, then x = (0 + 2)/5 = 2/5. If y = 8, then x = (8 + 2)/5 = 2, which, as you found, does give y = 8.

I don't know what you did in your work above, but you weren't evaluating this expression.
 
The answer is what you said here, and had said before (written correctly): x = (y + 2)/5.

As I said, solving for x means expressing x as an expression in terms of y, which is what this is.

With this solved formula, for ANY value of y, you can determine what x is. If y=0, then x = (0 + 2)/5 = 2/5. If y = 8, then x = (8 + 2)/5 = 2, which, as you found, does give y = 8.

I don't know what you did in your work above, but you weren't evaluating this expression.

Thanks again for your reply.
After a whole day spent working on solving, substituting an eliminating, I would not have understood what you have typed if I didn't do that today.
I still don't understand what you mean but I get the idea of it.

Expressions... aren't they used for substituting with another equation or line to get another expression? How can I get the coordinates of y and x for the graph? How can I stick x = (y + 2)/5 onto a graph?
 
Thanks again for your reply.
After a whole day spent working on solving, substituting an eliminating, I would not have understood what you have typed if I didn't do that today.
I still don't understand what you mean but I get the idea of it.

Expressions... aren't they used for substituting with another equation or line to get another expression? How can I get the coordinates of y and x for the graph? How can I stick x = (y + 2)/5 onto a graph?

An expression is any combination of numbers and variables that you can evaluate (find a value for). You can do all sorts of things with them; they are what algebra is all about.

I don't know what you mean by "stick onto a graph", but to graph the original equation y = 5x - 2, you can choose any value of x you want, find the value of y, and plot the point (x, y). The details are entirely up to you. You can do the same with the form x = (y + 2)/5, choosing any value of y and finding x, then plotting (x,y). These will, of course, be the same graph, since the equations are equivalent.
 
An expression is any combination of numbers and variables that you can evaluate (find a value for). You can do all sorts of things with them; they are what algebra is all about.

I don't know what you mean by "stick onto a graph", but to graph the original equation y = 5x - 2, you can choose any value of x you want, find the value of y, and plot the point (x, y). The details are entirely up to you. You can do the same with the form x = (y + 2)/5, choosing any value of y and finding x, then plotting (x,y). These will, of course, be the same graph, since the equations are equivalent.

Hmm.....
See this is the part when you start talking terms and I have no idea what you are saying. I would really like to see the equations on the graph.
When I see it, it makes sense to me, at the moment it all looks like just a bunch of numbers and letters that don't mean anything to me.
I tried that original equation on desmos to see what it meant, it's just a straight line.
Choose any value? What is a value? Are you meaning a coordinate on a Graph?
When you say finding y and finding x, I'm visualising the coordinates like the centre (h,k) which is (x,y). If it says "x - 2" I'm looking at the graph at -2 on the x axis. You know what? This will take a very long time trying to understand the theory, considering how long it is already taking! I am more interested in learning how to do the process. Knowing what to do when a situation/problem comes up and using that pattern or method to fix it. (It's like learning to do the dishes but not really knowing why you do them. It's just the process of doing it because you have to.)

Back to the main reason why I posted this thread.....

I am looking for the method of the equation x^2 = -28 (y - 8.5) to get my answer for the study.
What methods am I suppose to use? How do I do them? What is the answer??
That would help a lot
 
… What is a value? …
In the most general sense, a value is any Real-number. Here are some examples of Real values:

\(\displaystyle 1,024 \quad -55 \quad \dfrac{69}{7} \quad 11^{2} \quad 9.8\overline{571428} \quad \sqrt{43} \quad \pi \quad 6.022 \times 10^{-23} \quad 0.3333…\)


… When you say finding y and finding x, I'm visualising the coordinates like the centre (h,k) which is (x,y) …
Yes, your parabola is composed of an infinite number of points. (In fact, between any two points on the parabola that are as close together as you can imagine, there are still an infinite number of points between them! Individual points are infinitely microscopic, so they don't exist as real objects. They exist only in our mind; they're a very useful concept.)

Each point has an address which takes the form of an ordered pair of numbers (x,y) called coordinates. The symbols (h,k) represent the coordinates of the vertex point. The vertex of a parabola is the turning point of the curve. (You used 'centre', and it's true that the vertex is centered horizontally, but the vertex is not centered vertically. So, in the US, we prefer the name 'vertex'.)

Part of your goal is to determine the coordinates of the point on the parabola which is vertically above each end of the pool width. The y-coordinate of these points gives the height of the roof above the water -- at each end of the pool width.


… What methods am I suppose to use? How do I do them? …
These are good questions for you to ask your instructor. I don't even know what class you're taking, so I'm unable to answer these questions.

Also, you posted your exercise on the Calculus board, but it's not a calculus exercise. It's algebra. You mentioned that you've been doing implicit differentiation. That seems peculiar because (in my view) it is backwards to study calculus methods without first learning basic algebra.

Here's how I would do the exercise. Introduce a coordinate system on the given diagram, and label the known points with their coordinates.

poolRoof.JPG

Do the labeled points make sense? For examples: at the point where the roof meets the pool deck (to the right), the coordinates are (18,0). The x-coordinate measures horizontal distance from the vertical axis; so (18,0) is 18 meters away from the center of the pool's width (to the right). The point (-18,0) also measures horizontal distance, but in the opposite direction (to the left). The y-coordinate measures vertical distance from the horizontal axis; so (18,0) is 0 meters above the pool deck. The vertex point has coordinates (0,8.5). The x-coordinate is 0, so the point is zero meters away from the vertical axis. The y-coordinate is 8.5, so the point is 8.5 meters above the horizontal axis (i.e., the water's surface).

The horizontal distance from (-18,0) to (18,0) is 36 units. This matches the requirement that the width of the roof (at the pool deck) is no more than 36 meters.

The pool width is given as 18 meters. The vertical axis runs through the center of the width (cutting it in half), so the points at each end of the pool width must each be 9 meters away from the vertical axis. That's why the end points have coordinates (-9,0) and (9,0).

All parabolas that open either upward or downward (like yours) are graphs of what we call 'Quadratic Equations'. These equations may be expressed in various forms; I used this generic form:

y = a∙x^2 + b∙x + c

where symbols a,b,c each represent some fixed, Real number. For some arbitrary parabola, the values of a,b,c could each turn out to be any Real number, with one exception: symbol a cannot equal zero. If symbol a were equal to zero, then the term a∙x^2 would go away, and the resulting equation would be y=b∙x + c. But that's not quadratic; it's linear (its graph is a straight line, not a parabola).

If we know the coordinates of three points on a parabola, then we can find the values of a,b,c using an algebraic method known as "writing and solving a system of equations". It goes like this.

One at a time, substitute the x- and y-coordinates from a known point into the generic form and simplify (that is, replace all symbols x in the generic form with its Real value, and replace symbol y with its Real value.) Doing this gives a new quadratic equation containing only the symbols a,b,c (no symbol x and no symbol y).

Point #1: (-18,0)

0 = a∙(-18)^2 + b∙(-18) + c

Simplify:

0 = 324∙a - 18∙b + c

Point #2: (18,0)

0 = a∙(18)^2 + b∙(18) + c

Simplify:

0 = 324∙a + 18∙b + c

Point #3: (0,8.5)

8.5 = a∙(0)^2 + b∙(0) + c

Simplify:

8.5 = c

The three equations in red comprise our "system of equations". Did you notice? We just found the value of c.

Therefore, the Quadratic Equation we're looking for can now be written as:

y = a∙x^2 + b∙x + 8.5

We continue solving the rest of the system, to find the values of symbols a and b.

We can add equations, to get a new, valid equation. Let's add the first two (red) equations, in our system:

0 = 324∙a - 18∙b + 8.5
0 = 324∙a + 18∙b + 8.5
---------------------------
0 = 648∙a + 0∙b + 17

Simplify:

0 = 648∙a + 17

See what happened? -18∙b plus 18∙b combined to yield 0∙b because -18+18 is zero (opposites cancel each other out, when added). This is why I chose to add the first and second equations; I could see that symbol b would be eliminated from the result. Now we have an equation that contains only one symbol (a), so we can solve the equation and find the value of a.

We solve the equation 0 = 648∙a + 17 using two algebraic steps: (1) subtract 17 from each side, and (2) divide each side by 648.

(1) 0 - 17 = 648∙a + 17 - 17

Simplify:

-17 = 648∙a

(2) -17/648 = 648∙a/648

Simplify:

-17/648 = a

Now we know the value of a. We're getting closer; we need to find b:

y = -17/648∙x^2 + b∙x + 8.5

Let's go back to the second equation in our system: 0 = 324∙a + 18∙b + c

We have the values for symbols a and c, so we substitute them into this equation and simplify. That will yield a new equation containing only symbol b, so we can solve for b:

0 = 324∙(-17/648) + 18∙b + (8.5)

Simplify:

0 = -17/2 + 18∙b + 8.5

0 = -8.5 + 18∙b + 8.5

Opposites cancel when added, yes? Therefore, -8.5 + 8.5 is 0:

0 = 18∙b

We see that b must be zero.

We have found the Quadratic Equation whose graph is your parabola:

y = -17/648∙x^2 + 8.5

Let's write the fraction -17/648 in decimal form, and round it to four places:

y = -0.0262∙x^2 + 8.5

Okay. This equation is a formula for finding y, if we know x. Symbol y also represents the height of the parabola above the horizontal axis, at any given value of x on the horizontal axis.

In other words, we can use the formula to determine the height of the roof above each end of the pool width.

At the right end: x = 9

Substitute x = 9, in the formula:

y = -0.0262∙(9)^2 + 8.5

Simplify:

y = 6.3778

The coordinates of the point on the parabola vertically above the right end of the pool width are (9,6.3)

The y-coordinate represents the height of the roof, remember? Therefore, the roof is about 6.3 meters above the water, at the right end of the pool width.

Due to a concept known as symmetry (the right half of the parabola is a mirror image of the left half), we know that the y-coordinate at the left end of the pool width is the same.

We also know that between the ends of the pool width the roof is even higher than at the ends. Hence, the entire roof is always at least 6 meters above the surface of the water.

roofHeights.jpg

This is just one method, to approach the exercise.

By the way, these boards do not comprise an online classroom. Volunteer tutors at this forum do not generally have the time or motivation to type up lessons or step-by-step solutions because free algebra lessons already exist at many other sites on the Internet. For examples, you could take a free, beginning algebra course at coursera.org or at khanacademy.com. You could also check out lesson sites, like purplemath.com or mathisfun.com We are mainly here to help students who already have some idea of what they're supposed to be doing, but get stuck at some step. We guide them past that point, and then they continue on their own to finish (or to get stuck again, in which case they show their work and ask another specific question at that step). Please don't expect us to teach you algebra; I think you ought to enroll in a bonafide algebra course, before studying calculus (or whatever class you're currently taking).

Please read the forum's submission guidelines. Cheers :cool:

PS: Oops, I did not properly round the value 6.3778 to one decimal place, above. I wrote 6.3, but it should be 6.4

I don't want to edit my diagram, so just replace 6.3 with 6.4, everywhere it appears.

Additionally, I ought to have labeled the point where the x-axis and y-axis meet as (0,0). We call that point "the Origin".

PPS: In this exercise, we must assume that the pool is filled to the brim (i.e., the water's surface is level with the pool deck). This is not realistic, unless that's how they do it in New Zealand. In the real world, if we needed to know the true height of the roof above the water (at any location), then we would need to know how far below the pool deck the water surface lies.
 
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Otis! You have saved me!! Have been looking to find an answer for a whole week!
A week!!

All the websites you have mentioned have been visited and even more (socratic, cosmolearning, saylordotorg, math-only-math.com, wikihow, youtube, book for dummies, every google page that you can open). It's kind of difficult to understand considering some teach a different way to what I have been learning so don't want to go too far learning something that isn't part of my study. I was really trying to find a step to step guide that can branch into different methods depending on the problem an answer you are looking for.

All I can say now is THANK YOU THANK YOU THANK YOU!!

Now it is clear what to do, this guide can be very handy for future exercises!
I have noticed that one of the equations you have shown are very similar to the example I have used earlier in the posts below.
That means that where I was going with it was correct, but I was unaware that there is more to do afterwards.

Let's bring that up again:

The form used in the example is:
y= ax^2 + b
a = 18m (which is half of the base measurement)
b = 8.5m (vertex)

put it together...
y= 18^2 + 8.5

now if I work it out the way the example shows...
18^2 = 324
8.5/324 = 0.0262

to make sure this is correct...
-0.0262 x 18^2 = -8.5

so the equation is
y= -0.0262x^2 + 8.5

NOW.... I didn't know there was another part to it!
If you substitute the x point (9,0) (or any x point for that matter) in replace with the x in the equation. You get an answer for the height solving that equation!


I am not doing a class but a correspondence course. It's a self-study where they send you all the papers and you learn them yourself. At the end of the papers you take an exam and they assess you online. Part of the reason why I have been asking for guidance on this website, the study papers are all back to front and they are poorly written (I have already reported 5 mistakes so far). I'm starting to lose trust in learning something that is unreliable, I don't know how they mark their students but it's a little nerve racking. Next option would be to take classes!. It saves you about 600 bucks in not doing the class but then again it's about the quality. A personal lesson I guess...

Thank you thank you again Otis for you help. I am so relieved, it has been frustrating for all of us!
If there was a rating for each person on this website I would give you the top 5 stars! I am so grateful thank you.
 
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… All the websites you have mentioned have been visited …
Alright, but I was thinking not of visits but starting at the beginning of a structured presentation of basic algebra. Learning the concepts in beginning algebra will require a few months of watching lectures and working practice exercises. Continuing to learn the concepts in intermediate algebra will require a few additional months of study.

… don't want to go too far learning something that isn't part of my study.
What is your area of study?

If you're learning math, it's best to gain general knowledge in all the introductory topics. If you try to "specialize" (or your correspondence authors do), and seek out only information that pertains to some tasks at hand, you're really cheating yourself (or being cheated) of the big picture. Math instruction builds upon itself; the general skills you learn in one topic become tools in your toolkit for learning subsequent topics.


The form used in the example is:
y= ax^2 + b
a = 18m (which is half of the base measurement)
b = 8.5m (vertex)

put it together...
y= 18^2 + 8.5
I think Dr. Peterson already told you that example is not good for this exercise. Also, you did not "put it together" correctly.

If you're given a generic form of y = ax^2 + b and you're told that a=18 and b=8.5, then the substitution of those values looks like this:

y = 18x^2 + 8.5

That's the wrong equation, for the parabolic roof.

This is why it's better to use a structured approach; learn the general formula which works for any parabola that opens upward or downward:

y = ax^2 + bx + c

It seems like your materials might be trying to present some sort of "if the problem looks like this, then do it this way; or if the problem looks like that, then do it that way; or if the problem looks like a third situation, then do it a third way" sort of system. Creating extra rules/shortcuts to cover different situations is unnecessary, when there's a general approach that works in all situations. Why learn different rules for different situations, when there's a general approach that works in all cases?


I am not doing a class but a correspondence course. It's a self-study … the [materials] are poorly written … [but taking this course] saves you about 600 bucks …
The online algebra courses and the review sites that I mentioned are FREE. The only expense is your time. It takes several months of study and practice, to gain a strong, complete foundation in algebra -- a foundation that's an essential pre-requisite for studying calculus. (A complete introduction to calculus requires also trigonometry and a handful of pre-calculus topics.) Again, I don't know why you're studying math, but if your goals require a lot of it, you'll save time in the long run by completing a structured course of study prepared by professional mathematicians and educators. :cool:
 
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