Linear equation and parabola?

[Frosted_Knickers]

Hi everyone,
Hope all is well. I am currently doing a maths assignment based on linear functions and relationships. I have come across a question which is driving me a bit nuts, there is so much to learn in this topic alone.
For the following algebraic expressions, I need to find out the value of x and y, and place them on a graph. After this, I have to verify my answers under another section.

5x-2y=3
y-2x^2=-5 (the 2x^2 is supposed to be 2x squared, but I can't change the format on here. Sorry!)

Now, upon completing the first equation (5x-2y=3), I discovered that it is a linear equation.
For x I got (2y+3) ÷ 5, and for y I got -5x-3 ÷ 2. I drew a graph myself, assuming x or y is 0, for instance (x and y intercepts). I double checked I got this correct using Desmos and everything is good so far.

Now here is where I get stuck. For the second equation y-2x^2=-5, I know this is not a linear function. There is a squared variable (x), and so I think it is a parabola? I have watched several khan academy videos regarding this, particularly vertex and quadratic functions. Correct me if I am wrong please, but I don't think this equation can fit under the quadratic expression? y= ax^2+bx+c [edited], however I heard it won't work if the ax^2 is a negative. One of the websites suggested the vertex formula (y= a(x-h)^2+k), however I am not sure how this would work if I haven't even graphed my numbers yet? I am still trying to figure out the algebraic values of both x and y. A final option was to swap numbers from the equation to isolate both x and y. I tried this, but got a slightly different answer, and the sources never elaborated how they got the final answer (one of those maths scanning apps).

So, out of desperation and curiosity, does anyone here have any thoughts on what formula to use? I don't really know a great deal about linear functions, so I am going to learn a lot!
Thanks guys!

 

[Dr.Peterson]

Now here is where I get stuck. For the second equation y-2x^2=-5, I know this is not a linear function. There is a squared variable (x), and so I think it is a parabola? I have watched several khan academy videos regarding this, particularly vertex and quadratic functions. Correct me if I am wrong please, but I don't think this equation can fit under the quadratic expression? y= ax^2+bx+c, however I heard it won't work if the ax^2 is a negative. One of the websites suggested the vertex formula (y= a(x-h)^2+k), however I am not sure how this would work if I haven't even graphed my numbers yet? I am still trying to figure out the algebraic values of both x and y. A final option was to swap numbers from the equation to isolate both x and y. I tried this, but got a slightly different answer, and the sources never elaborated how they got the final answer (one of those maths scanning apps).

So, out of desperation and curiosity, does anyone here have any thoughts on what formula to use? I don't really know a great deal about linear functions, so I am going to learn a lot!
Thanks guys!

When you solve the second equation for y, you will get y = 2x^2 - 5. The coefficient of x^2 is not negative!

But that doesn't even matter; if it had been y = -2x^2 - 5, it would just be a parabola flipped over, opening downward. It "works" fine.

You can use the vertex form by completing the square, but you don't have to unless you were told to. If you are just starting graphing, as it appears, then just plot some points (x, y) and see what you get.

Can you show us the exact wording of the entire problem you are working on, including instructions? Also, what topics are covered in the assignment? That should give us a better idea what to suggest for you.

 

[Frosted_Knickers]

Hi Dr Peterson, thanks for your reply! I see what you mean in this regard when values are swapped. I guess my question would be, now that 2x^2 can be formed into a positive, would this actually be a quadratic function (ax^2+bx+c)? I know this is to discover what y is, however, I don't see any bx value of the equation y=-2x^2-5? I only see the ax^2 value and c? Unless there is a trick in which I assume bx equals zero, or do more substitutions. I dont think we have even quite learned the "closing the square" formula yet, although I will definitely investigate if all else fails.

Not a problem, there are three parts linked to the same question. The wording is as follows:

a) Solve the following simultaneous equations algebraically to find x and y.
5x-2y=3
y-2x^2=-5

b) Graph both equations on the same set of axes

c)Using the graph in (b), verify your solutions to (a).

I wanted to mention that I managed to correctly graph both equations by some miracle. I wanted to double check these on desmos, as I have heard it is a good graphing website. I'm also on the khan academy website, just getting my head around the linear functions/relationships and each sub-category. It's quite a big topic to cover!

Thank you kindly for your help, I know any help I receive will go a very long way. I greatly appreciate it!

 

[Deleted member 4993]

Hi Dr Peterson, thanks for your reply! I see what you mean in this regard when values are swapped. I guess my question would be, now that 2x^2 can be formed into a positive, would this actually be a quadratic function (ax^2+bx+c)? I know this is to discover what y is, however, I don't see any bx value of the equation y=-2x^2-5? I only see the ax^2 value and c? Unless there is a trick in which I assume bx equals zero, or do more substitutions. I dont think we have even quite learned the "closing the square" formula yet, although I will definitely investigate if all else fails.

Not a problem, there are three parts linked to the same question. The wording is as follows:

a) Solve the following simultaneous equations algebraically to find x and y.
5x-2y=3
y-2x^2=-5

b) Graph both equations on the same set of axes

c)Using the graph in (b), verify your solutions to (a).

I wanted to mention that I managed to correctly graph both equations by some miracle. I wanted to double check these on desmos, as I have heard it is a good graphing website. I'm also on the khan academy website, just getting my head around the linear functions/relationships and each sub-category. It's quite a big topic to cover!

Thank you kindly for your help, I know any help I receive will go a very long way. I greatly appreciate it!

y₁ = ax² + bx + c .......................................(1)

and

y₂ = ax² + c .......................................(2)

and

y₃ = ax² .......................................(3)

and

y<sub>4/SUB] = ax² + bx .......................................(4)

All four equations above are quadratic equation, as long as

a \(\displaystyle \ne \) 0

The signs (positive or negative) of the coefficients (a or b or c) do not matter.</sub>

 

[Dr.Peterson]

Hi Dr Peterson, thanks for your reply! I see what you mean in this regard when values are swapped. I guess my question would be, now that 2x^2 can be formed into a positive, would this actually be a quadratic function (ax^2+bx+c)? I know this is to discover what y is, however, I don't see any bx value of the equation y=-2x^2-5? I only see the ax^2 value and c? Unless there is a trick in which I assume bx equals zero, or do more substitutions. I dont think we have even quite learned the "closing the square" formula yet, although I will definitely investigate if all else fails.

I'm not sure what you mean by "swapped". I don't think I swapped anything. I did solve for y, which moves values around.

Again, positive or negative doesn't matter. Solving for y, which you do to graph an equation (by writing it as a function) resulted in y = 2x^2 - 5, but it would still have been a quadratic function if it were y = -2x^2 - 5.

When you are introduced to quadratic equations, you should be taught that a, b, and c in the formula can be any numbers at all, even zero -- except that if a were zero, it wouldn't be called a quadratic equation, but linear. When there is no "x" term, it just means that b=0:

y = -2x^2 - 5 is the same as y = -2x^2 +0x - 5, so a=-2, b=0, and c=-5.​

I don't think you've yet told us what you have learned; clearly there is a lot that you have not, and knowing what we can expect you to know will be important.

There are three parts linked to the same question. The wording is as follows:

a) Solve the following simultaneous equations algebraically to find x and y.​

5x-2y=3​

y-2x^2=-5​

​

b) Graph both equations on the same set of axes​

​

c)Using the graph in (b), verify your solutions to (a).​

Shall I assume that you did part (a) successfully, which does not involve graphing? What you have been asking about is part (b). Your initial description of the problem suggested that you may be misinterpreting it. No part says, "For the following algebraic expressions, ... find out the value of x and y, and place them on a graph."

Again, it will be very helpful to know your background, as the wording of the problem implies that you are expected to have plenty of experience graphing these kinds of equations already, as that is just mentioned in passing. Also, this is not, as you said, "a maths assignment based on linear functions and relationships"; it is about non-linear systems of equations.