Beans06 Posted October 31, 2015 Share Posted October 31, 2015 How do u find the minimum parabla that has x-intercepts of -2 and 4 and passes through the point (3,-15) ? I have that the answer is -27 but I don't get how to get that answer can some please help ? Link to comment Share on other sites More sharing options...
studiot Posted October 31, 2015 Share Posted October 31, 2015 How do u find the minimum parabla that has x-intercepts of -2 and 4 and passes through the point (3,-15) ? I have that the answer is -27 but I don't get how to get that answer can some please help ? Have you heard of simultaneous equations? What is the general equation between x and y for the parabola? How many constants does it have? and at how many points do you know both x and y? Link to comment Share on other sites More sharing options...
Beans06 Posted October 31, 2015 Author Share Posted October 31, 2015 No I never heard of that but those are the only points that the question gave me Link to comment Share on other sites More sharing options...
studiot Posted October 31, 2015 Share Posted October 31, 2015 (edited) So what is the equation for a parabola? Alternatively, what do you know about parabolas? Note we do not do your homework for you, just help you find the way. I do not know what maths you should know, but if you are studying quadratic equations (parabolas) then you should have met simultaneous equations. Can you not solve for instance 3x + 2y = 7 3x - 2y = -1 These are simultaneous equations. Incidentally I agree that the minimum is at x=1, y=-27. Edited October 31, 2015 by studiot Link to comment Share on other sites More sharing options...
Beans06 Posted October 31, 2015 Author Share Posted October 31, 2015 (edited) That's what I was wondering if u could help me get the equation I don't get what I have to do next ? I know how to solve that equation x= 1 and y =2 I was wondering if you could help me understand how to get the minimum parabola with out having a equation to start with ? I know about a parabola but only when I have an equation then I can find the maximum or minimum but with out and equation it's hard to find that or I just don't understand how to find that . Edited October 31, 2015 by Beans06 Link to comment Share on other sites More sharing options...
Beans06 Posted October 31, 2015 Author Share Posted October 31, 2015 how do you find the minimum of a parabola that has the x- intercepts -2 and 4 and goes through the point (3,-15)? I don't get how to get the minimum if I don't have a starting equation to work with can someone please help me on this ? Link to comment Share on other sites More sharing options...
Essay Posted October 31, 2015 Share Posted October 31, 2015 You should be able to write your "starting equation," or 'reconstruct' it, based upon the information that you're given. Link to comment Share on other sites More sharing options...
moth Posted October 31, 2015 Share Posted October 31, 2015 Khan academy has good videos about algebra. Have you tried looking at any of those? Link to comment Share on other sites More sharing options...
studiot Posted October 31, 2015 Share Posted October 31, 2015 (edited) So if I wrote ax + by = 7 cx + dy = -1 Would you understand it and what would you need to solve these two equations? If you don't understand I will explain in more detail. This will lead to the more complicated example of your parabola. Edited October 31, 2015 by studiot Link to comment Share on other sites More sharing options...
pavelcherepan Posted October 31, 2015 Share Posted October 31, 2015 Beans06, the very generalised formula for a parabola is ax2+bx+z=y, right? So, now your available data gives you three data points, so for each of those you can create an own version of the equation, for example for the -2 intercept you'll have: a(-2)2+(-2)b+z=0 Do the same for intersection at 4 and (3, -15) and you'll have a set of 3 equations with 3 unknowns. Now you'll just need to solve those. Link to comment Share on other sites More sharing options...
fiveworlds Posted October 31, 2015 Share Posted October 31, 2015 Beans06 what you need to do is find any parabola to do this take the two x intercepts and any point. Say x=3 and x=7 with the point (22,156) Then change the sign of the intercepts and write it like this y=a(x-3)(x-7). Then sub your point into the equation given. This allows you to calculate a which will give you the equation of a parabola. Link to comment Share on other sites More sharing options...
studiot Posted October 31, 2015 Share Posted October 31, 2015 Now you'll just need to solve those. Actually you also need some calculus in theory. Beans06 what you need to do is find any parabola to do this take the two x intercepts and any point. Say x=3 and x=7 with the point (22,156) Then change the sign of the intercepts and write it like this y=a(x-3)(x-7). Then sub your point into the equation given. This allows you to calculate a which will give you the equation of a parabola. As pavel said, there are three constants to find, not one. Link to comment Share on other sites More sharing options...
Beans06 Posted October 31, 2015 Author Share Posted October 31, 2015 (edited) That's what I did y=a(x+2)(x-4) but it didnt give me that minimum of -27? Edited October 31, 2015 by Beans06 Link to comment Share on other sites More sharing options...
studiot Posted October 31, 2015 Share Posted October 31, 2015 Let's just keep going with the simultaneous equations in post9 Then we can use our information in pavels equation from post 10 to get the correct answer. Link to comment Share on other sites More sharing options...
Beans06 Posted October 31, 2015 Author Share Posted October 31, 2015 Okay can u please explain it in more detail? I kind of understand it but not fully Link to comment Share on other sites More sharing options...
studiot Posted October 31, 2015 Share Posted October 31, 2015 OK so let's provide some motivation. Pavel has offered the equation of a parabola ax2 + bx + z = y Where a, b and z are constants. This is also known as a quadratic equation. But you are not asked to solve the quadratic, you are given the values of x and y at three points on the curve. x = -2 , y = 0 and x = 4, y = 0 and x = 3, y = -15 So can you write down what happens if you substitute these values into the quadratic equation of the parabola? What do you get? Link to comment Share on other sites More sharing options...
Beans06 Posted November 1, 2015 Author Share Posted November 1, 2015 Okay so far I get it. what I don't get is what do I do with the z because by substituting I got : 4a-2b+z=0 16a+4b+z=0 9a+3b+z=-15 Link to comment Share on other sites More sharing options...
pavelcherepan Posted November 1, 2015 Share Posted November 1, 2015 (edited) Okay can u please explain it in more detail? OK. So as I said before the standard form for the equation of a parabola is ax2+bx+c=y. In order for you to calculate the minimum of the function you need to find all the three constants a,b, and c. Obviously, to find three unknowns you need a minimum of three equations. Thankfully, you have 3 data points (-2,0); (4,0) and (3,-15). Now using these value you can construct 3 equations. For the first 2 points you simply just put x-value of the intercept instead of x in the formula and y will be equal to 0. For the last one everything is the same, except for y will be equal to -15. After all this is done, you'll have a set of three equations. Take one of them, solve it for a, for example, then put the resulting value for a into the second equation. Now you'd only be dealing with 2 unknowns. Solve this equation for one of the two remaining unknowns and then plot the result into the third equation and then you'll be able to find all three constants. Now that you've got yourself the proper formula for the parabola you can find the minimum in two ways. If you already have done differentials then differentiate the equation for x and equate the resulting equation to 0, since at the minimum/maximum point of a parabola the rate of change of the function is 0. Or else you can just look up the standard formula for the minimum of a parabola. Google will help with that one. Edited November 1, 2015 by pavelcherepan Link to comment Share on other sites More sharing options...
studiot Posted November 1, 2015 Share Posted November 1, 2015 Okay so far I get it. what I don't get is what do I do with the z because by substituting I got : 4a-2b+z=0 16a+4b+z=0 9a+3b+z=-15 Yeay you are getting there. You get three equations connecting a, b and z . But because they refer to the same parabola they are the same a , b and z in each equation. We say the equations are simultaneous. So you can solve these three equations to find a, b and z, as you did back in post 5 In order to help a little bit I will tell you that a = 3 is a good guess, or you can just go ahead and find all three. What do you make a, b and z? So when you have found a,b and z you can write down the equation of your particular parabola, that passes through the given points. So you can find any point on it if you know x or y, you can calculate the other. The point you want to find is the minimum. Now I asked you back in post4 what you know about the parabola as you have two ways to do this. You can either use the calculus to differentiate the equation and se this equal to zero to find the minimum if you know how to do this. Or you can use the geometric properties of the parabola (in this case the symmetries) to find the axis and thus the vertex, which is the geometrical name for the minimum. So we await your next input. Link to comment Share on other sites More sharing options...
fiveworlds Posted November 1, 2015 Share Posted November 1, 2015 As pavel said, there are three constants to find, not one Using the method I was using I found the parabola 3x^2-6x-24 Which gives @x=-2 (12)+12-24=0 true @x=4 (48)-24-24=0 true @x=3 (27)-18-24=-15 true Link to comment Share on other sites More sharing options...
studiot Posted November 1, 2015 Share Posted November 1, 2015 Fiveworlds, it would be interesting to know how you continued your analysis to find the minimum. There are several non mainstream methods that could be used in attacking this whole problem. Yours sounds equivalent to moving the parabola to a new position relative to axes. Link to comment Share on other sites More sharing options...
Nicholas Kang Posted November 1, 2015 Share Posted November 1, 2015 (edited) I shall give it a try. Let's deal with the equations first. You had followed what studiot said. So, solve the three equations. (In Malaysia, we are taught that general quadratic equation is ax2+bx+c=0; we don't use z in place of c) 4a-2b+c=0--EQ1 16a+4b+c=0--EQ2 9a+3b+c=0--EQ3 EQ1 x2-- 8a-4b+2c=0 4b=8a+2c Substitute 4b=8a+2c into EQ1 You get: 16a+8a+2c=0 24a+3c=0 3c=-24a c=-8a Substitute c=-8a into EQ1 You get: 4a-2b-8a=0 -2b=-4a b=-2a Substitute b=-2a and c=-8a into EQ3 9a+3(-2a)+(-8a)=-15 9a-6a-8a=-15 -3a=-15 a=-3 Substitute a=-3 into b=-2a and c=-8a You get a=-3, b=-6, c=-24 Quadratic equation is 3x2-6x-24=0 To get the minimum value, I invoke the method of completing the square (This was taught in Form 4, 16-year-old syllabus in Malaysia) To complete the square, first you have to understand how it works. Let's say we have quadratic equations: (x-1)(x-1)=0 (x-2)(x-2)=0 (x-3)(x-3)=0 So, x2-2x+1=0 x2-4x+4=0 x2-6x+9=0 Notice the pattern of the above 3 equations. You can conclude: x2-2x=(x-1)2-12 x2-4x=(x-2)2-22 x2-6x=(x-3)2-32 So, x2-nx=(x-n/2)2-(n/2)2 Now, 3x2-6x-24=0 3(x2-2x)-24=0 3[(x-1)2-12]-24=0 3(x-1)2-3-24=0 3(x-1)2-27=0 After completing the square, you will obtain a formula that looks like a(x-s)2+v=0, where s represents the axis of symmetry and v represents the minimum/maximum value of the function/equation In this case x-1=0 thus axis of symmetry is x=1 whereas -27 is the minimum value of the equation. The answer is -27. (I posted a link in case you need some references. The link below directs you to a Malaysian Form 4/16-year-old Malaysia Education Certificate Syllabus-based Additional Mathematics website. Look for Chapter 2(quadratic equations) and 3(quadratic functions) for details.) http://spmaddmaths.onlinetuition.com.my/ Edited November 1, 2015 by Nicholas Kang Link to comment Share on other sites More sharing options...
DimaMazin Posted November 4, 2015 Share Posted November 4, 2015 I shall give it a try. Let's deal with the equations first. You had followed what studiot said. So, solve the three equations. (In Malaysia, we are taught that general quadratic equation is ax2+bx+c=0; we don't use z in place of c) 4a-2b+c=0--EQ1 16a+4b+c=0--EQ2 9a+3b+c=0--EQ3 EQ1 x2-- 8a-4b+2c=0 4b=8a+2c Substitute 4b=8a+2c into EQ1 You get: 16a+8a+2c=0 24a+3c=0 3c=-24a c=-8a Substitute c=-8a into EQ1 You get: 4a-2b-8a=0 -2b=-4a b=-2a Substitute b=-2a and c=-8a into EQ3 9a+3(-2a)+(-8a)=-15 9a-6a-8a=-15 -3a=-15 a=-3 Substitute a=-3 into b=-2a and c=-8a You get a=-3, b=-6, c=-24 Quadratic equation is 3x2-6x-24=0 To get the minimum value, I invoke the method of completing the square (This was taught in Form 4, 16-year-old syllabus in Malaysia) To complete the square, first you have to understand how it works. Let's say we have quadratic equations: (x-1)(x-1)=0 (x-2)(x-2)=0 (x-3)(x-3)=0 So, x2-2x+1=0 x2-4x+4=0 x2-6x+9=0 Notice the pattern of the above 3 equations. You can conclude: x2-2x=(x-1)2-12 x2-4x=(x-2)2-22 x2-6x=(x-3)2-32 So, x2-nx=(x-n/2)2-(n/2)2 Now, 3x2-6x-24=0 3(x2-2x)-24=0 3[(x-1)2-12]-24=0 3(x-1)2-3-24=0 3(x-1)2-27=0 After completing the square, you will obtain a formula that looks like a(x-s)2+v=0, where s represents the axis of symmetry and v represents the minimum/maximum value of the function/equation In this case x-1=0 thus axis of symmetry is x=1 whereas -27 is the minimum value of the equation. The answer is -27. (I posted a link in case you need some references. The link below directs you to a Malaysian Form 4/16-year-old Malaysia Education Certificate Syllabus-based Additional Mathematics website. Look for Chapter 2(quadratic equations) and 3(quadratic functions) for details.) http://spmaddmaths.onlinetuition.com.my/ Some features of a parabola Coordinates of the vertexThe x-coordinate at the vertex can be found by completing the square to put the equation in vertex form, or by differentiating the original equation, setting the resulting equal to zero (a critical point), and solving for . Both methods yield . Substituting this into the original equation yields https://en.wikipedia.org/wiki/Parabola a=3 not -3 y= -24 - (-6)2/(4 * 3) 1 Link to comment Share on other sites More sharing options...
studiot Posted November 4, 2015 Share Posted November 4, 2015 (edited) Thank you for your summary, DimaMazin +1 It should be noted that your pink parabola from Wiki is not the one we are discussing, although it shows the important features. You do not need to use calculus or algebra to obtain the vertex x coordinate, which is what I meant by saying to use the geometric properties of a parabola. We are assuming an upright parabola, and the coefficient a is positive which means it is concave up, so has a minimum which we want. The x axis is a chord of this parabola, perpendicular to the axis, and any such chord is bisected by a vertical line through the vertex. So the vertex is halfway between the intercepts on the x axis (-2 and +4) ie through x = +1. Substituting this x value into the parabola equation already found does indeed yield y = -27. Edited November 4, 2015 by studiot 1 Link to comment Share on other sites More sharing options...
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