I can’t figure out what im Doing wrong with this question.
The correct answer is suppose to be 104.5 but when I try to do the x intercept I get numbers like 17. and when I do it as 800-0.1s…etc it comes out as something like 722.
Any ideas? maybe im not using correct method on calc.
i put it into y1 and y2 then graph, put into window where i can see the intersection of x then i do 2nd trace and intersect. when i zoom out of graph it appears to be a parabola. the intersect want two different curves but there is only 1?
I did it the exact same way. I check the numbers like 5 times. I feel like there is something fucked up about my calculation. Ive had problems with graphs for a couple days. It never comes out right. Of course my final is in a couple hours.
I dunno what exactly you are doing wrong, but the answer is given by the function f(x)=800-(.1x^2-3x+22) Most likely you were just forgetting the parenthesis around the y2. This gives you a parabola opening downward crossing the x-axis around 104.5
here’s a picture, its large. The inserted function is top left, and I switched the tic marks to increments of 104.5 to show the x-intercept. Good luck on your final
[quote]coolnatedawg wrote:
Man, just use the quadriatic equation.
Set the formula as 0=.1s^2 - 3s - 778 and throw that into the quadriatic equation.
I returned s= 104.5. Do I need to show my work?
(Note: It’s probably easier to do in a calculator but I wanted to use my brain before the zombies kill me and it serves me no purpose anymore)[/quote]
While that is the right answer its the wrong way to do it. By doing it like that you give yourself a parabola that opens up. Had the question also asked about the vertex of the parabola you’d be wrong. Remember that its y1-y2=0, not y1=y2 to get the formula
[quote]coolnatedawg wrote:
Man, just use the quadriatic equation.
Set the formula as 0=.1s^2 - 3s - 778 and throw that into the quadriatic equation.
I returned s= 104.5. Do I need to show my work?
(Note: It’s probably easier to do in a calculator but I wanted to use my brain before the zombies kill me and it serves me no purpose anymore)[/quote]
While that is the right answer its the wrong way to do it. By doing it like that you give yourself a parabola that opens up. Had the question also asked about the vertex of the parabola you’d be wrong. Remember that its y1-y2=0, not y1=y2 to get the formula[/quote]
I was just solving the part above the break in the question. My answer was either this or a negative number. I did not have to use y1 or y2. I simply used s=(b^2 +/- sqrt(b^2-4ac))/2a.
Right? Or do I not understand what you are saying? It’s been a while…
[quote]coolnatedawg wrote:
Man, just use the quadriatic equation.
Set the formula as 0=.1s^2 - 3s - 778 and throw that into the quadriatic equation.
I returned s= 104.5. Do I need to show my work?
(Note: It’s probably easier to do in a calculator but I wanted to use my brain before the zombies kill me and it serves me no purpose anymore)[/quote]
While that is the right answer its the wrong way to do it. By doing it like that you give yourself a parabola that opens up. Had the question also asked about the vertex of the parabola you’d be wrong. Remember that its y1-y2=0, not y1=y2 to get the formula[/quote]
I was just solving the part above the break in the question. My answer was either this or a negative number. I did not have to use y1 or y2. I simply used s=(b^2 +/- sqrt(b^2-4ac))/2a.
Right? Or do I not understand what you are saying? It’s been a while…
[/quote]
I think what he means is that while you’re technically finding the correct answer for s, you’re not doing it in the manner the question asks and thus wouldn’t get full marks since usually there’s more emphasis places on the process than the answer itself in these types of questions.
[quote]coolnatedawg wrote:
Man, just use the quadriatic equation.
Set the formula as 0=.1s^2 - 3s - 778 and throw that into the quadriatic equation.
I returned s= 104.5. Do I need to show my work?
(Note: It’s probably easier to do in a calculator but I wanted to use my brain before the zombies kill me and it serves me no purpose anymore)[/quote]
While that is the right answer its the wrong way to do it. By doing it like that you give yourself a parabola that opens up. Had the question also asked about the vertex of the parabola you’d be wrong. Remember that its y1-y2=0, not y1=y2 to get the formula[/quote]
I don’t understand what you are doing. I thought the question just asked to solve for s. I got (15,-0.5) for the vertex of the parabola. I used the “vertex form” y = a(x â?? h)^2 + k.
[quote]coolnatedawg wrote:
Man, just use the quadriatic equation.
Set the formula as 0=.1s^2 - 3s - 778 and throw that into the quadriatic equation.
I returned s= 104.5. Do I need to show my work?
(Note: It’s probably easier to do in a calculator but I wanted to use my brain before the zombies kill me and it serves me no purpose anymore)[/quote]
While that is the right answer its the wrong way to do it. By doing it like that you give yourself a parabola that opens up. Had the question also asked about the vertex of the parabola you’d be wrong. Remember that its y1-y2=0, not y1=y2 to get the formula[/quote]
I was just solving the part above the break in the question. My answer was either this or a negative number. I did not have to use y1 or y2. I simply used s=(b^2 +/- sqrt(b^2-4ac))/2a.
Right? Or do I not understand what you are saying? It’s been a while…
[/quote]
I think what he means is that while you’re technically finding the correct answer for s, you’re not doing it in the manner the question asks and thus wouldn’t get full marks since usually there’s more emphasis places on the process than the answer itself in these types of questions.[/quote]
Is there another method to solve it without using the quadratic equation?
[quote]coolnatedawg wrote:
Man, just use the quadriatic equation.
Set the formula as 0=.1s^2 - 3s - 778 and throw that into the quadriatic equation.
I returned s= 104.5. Do I need to show my work?
(Note: It’s probably easier to do in a calculator but I wanted to use my brain before the zombies kill me and it serves me no purpose anymore)[/quote]
While that is the right answer its the wrong way to do it. By doing it like that you give yourself a parabola that opens up. Had the question also asked about the vertex of the parabola you’d be wrong. Remember that its y1-y2=0, not y1=y2 to get the formula[/quote]
I was just solving the part above the break in the question. My answer was either this or a negative number. I did not have to use y1 or y2. I simply used s=(b^2 +/- sqrt(b^2-4ac))/2a.
Right? Or do I not understand what you are saying? It’s been a while…
[/quote]
I think what he means is that while you’re technically finding the correct answer for s, you’re not doing it in the manner the question asks and thus wouldn’t get full marks since usually there’s more emphasis places on the process than the answer itself in these types of questions.[/quote]
Well then you think that would be an argument more in my favor then since a calculator wont show any work.
Also, the way I read the question, it tells you to solve the problem. It then says “to use the calculator to solve”. I interpret that as “if you are using the calc method, here is a hint.”
What teacher prefers you use a calculator over solving it by hand?
[quote]coolnatedawg wrote:
Man, just use the quadriatic equation.
Set the formula as 0=.1s^2 - 3s - 778 and throw that into the quadriatic equation.
I returned s= 104.5. Do I need to show my work?
(Note: It’s probably easier to do in a calculator but I wanted to use my brain before the zombies kill me and it serves me no purpose anymore)[/quote]
While that is the right answer its the wrong way to do it. By doing it like that you give yourself a parabola that opens up. Had the question also asked about the vertex of the parabola you’d be wrong. Remember that its y1-y2=0, not y1=y2 to get the formula[/quote]
I was just solving the part above the break in the question. My answer was either this or a negative number. I did not have to use y1 or y2. I simply used s=(b^2 +/- sqrt(b^2-4ac))/2a.
Right? Or do I not understand what you are saying? It’s been a while…
[/quote]
I think what he means is that while you’re technically finding the correct answer for s, you’re not doing it in the manner the question asks and thus wouldn’t get full marks since usually there’s more emphasis places on the process than the answer itself in these types of questions.[/quote]
Is there another method to solve it without using the quadratic equation?[/quote]
Yes, the one outlined. Of course we can simply directly solve for s using the quadratic formula, or solve for the intersection points between the curves y1=800 and y2=0.1s^2 - 3s + 22 as I did graphically in my first reply, or we can do it the technically correct way the question asks which is how Tyrant did it, and solve the equation f(x) = y1 - y2 = 800 - (0.1s^2 - 3s + 22).
If you’re still confused look at the graph in my reply, and look at the graph in Tyrant’s, his opens down and the solution is found via the X-INTERCEPT, where mine opens up and the solution is found via the CURVE intersect.
