Algebra multi-part question

[amathproblemthatneedsolve]

1. The question is as follows p,oint (1, 5) is on the curve: y = ax^2 + bx + c. This point gives the linear equation: 5 = a + b + c. A second point on the curve, (2, 10) gives the linear equation 10=4a+2b+c


A kid called Mike thinks that the point (2, 19) is also on the curve.

a. Use the point (2, 19) to write the third equation.

19=2a+b+c ?

b. Attempt to solve this system of three linear equations.

If my above equation is correct then it should be a=14,b=-37,c=28. Meaning the system is consistent with one solution.



2. Mike realizes that he has made an error and that the third point should be (3, 19) not (2, 19) as in question 1 above. • Rewrite the third equation and solve the new system of three linear equations. • Write down the quadratic function.

by using 19=3a+b+c I get a=7,b=-16,c=14 Write down the quadratic function. ?

3. Suppose that the third point is written as (3, t). Find all values for t that will change the quadratic function y = ax^2 + bx + c into a linear function

 

[Steven G]

1a)
You plugged in x=2 perfectly the first time but messed up in the 2nd attempt.
If (2,10) is on the curve then 10=4a+2b+c.
Now if (2,19) is on the curve then 19=4a+2b+c NOT 19=2a+b+c
Obviously you do not know what a function is or what properties that a quadratic equation have. In a function you can NOT have two different outputs (like 10 and 19) for one input (like 2).
Right away you should see that 4a+2b+c can't equal both 10 and 19.

2) 19=3a+b+c is not correct. Please try plugging in 3 for x in the equation y = ax^2 + bx + c.
Right or wrong you claim you found the values for a, b and c. Just plug those values into y = ax^2 + bx + c

 

[amathproblemthatneedsolve]

1a) Okay I see what you're saying! because plugging 2 into x^2 = 2^2=4 etc ect. Yes "Right away you should see that 4a+2b+c can't equal both 10 and 19." I do see this so this means no solution exists. Are equations 2 and 3 parallel or something?

2)
19=9a+3b+c solving for this gives me a=2,b=-1,c=4

Write down the quadratic function. ? does this just mean y = ax^2 + bx + c, y=2x^2-1x+c or do I use my a known equation like 19=3(2)-1+4

 

[Steven G]

1a) Okay I see what you're saying! because plugging 2 into x^2 = 2^2=4 etc ect. Yes "Right away you should see that 4a+2b+c can't equal both 10 and 19." I do see this so this means no solution exists. Are equations 2 and 3 parallel or something?

2)
19=9a+3b+c solving for this gives me a=2,b=-1,c=4

Write down the quadratic function. ? does this just mean y = ax^2 + bx + c, y=2x^2-1x+c or do I use my a known equation like 19=3(2)-1+4

You need to get a better understanding of what " = " means. When you write/say/think a = 2 that means that you can replace 2 for a or a for 2. It does NOT mean to replace 2 for x. Why? Because you did not conclude that x=2. Besides how can x=2 when you plugged in 1 and 3 for x. x is a variable which means it can take on many values. a, b and c are unknown but are constants. That is they are all just one number (maybe to be found) and do not change.

The fact that 4a+2b+c can't equal both 10 and 19 never needed to be noticed in this problem. What needed to be noticed was that a quadratic equation can never have both (2,10) and (2,19) as points satisfying the equation y = ax^2 + bx + c. However if you missed that one then the fact that 4a+2b+c needed to equal both 10 and 19 should have caused you concerns.

 

[Steven G]

Now what about part c?

What would have to true about some or all of the values for a, b and c so that y=ax^2 + bx + c is a line?

 

[Steven G]

1a) Okay I see what you're saying! because plugging 2 into x^2 = 2^2=4 etc ect. Yes "Right away you should see that 4a+2b+c can't equal both 10 and 19." I do see this so this means no solution exists. Are equations 2 and 3 parallel or something?

2)
19=9a+3b+c solving for this gives me a=2,b=-1,c=4

Write down the quadratic function. ? does this just mean y = ax^2 + bx + c, y=2x^2-1x+c or do I use my a known equation like 19=3(2)-1+4

I was initially confused about your equation 19=3(2)-1+4, but after reading it again I now have absolutely no idea what you mean. I'm afraid to read it again! It seems that you are trying to substitute the values for a, b and c into 19=9a+3b+c (??) but did a terrible job at it. if that was your goal, then it should read as 19 = 9(2) + 3(-1) + 4. Is that equation true? How about 19=3(2)-1+4, is that true? You need to check if what you write is valid. I am certain that you know what 3(2)-1+4 equals and it is not 19. So why do you write things like that? You don't check your own work.

If you do not check your own work then who will? The answer is your teacher who grades your exams and possibly your homework. Sure they will tell you where you made your silly mistakes but they also will deduct points off your final score. Check your own work!