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Math Help - Help me with graphing parabolas

  1. #16
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    Quote Originally Posted by Jhevon View Post
    If you don't want to use decimals, then use the exact values. Using integers would be considered wrong. For instance, if you used integers you would have the x-intercepts being (-4,0) and (0,0). The (0,0) says the graph goes through the origin, which we know is not the case. Don't worry about how it looks, just make sure your calculations are correct. If you get roots in your answer for an intercept, thats fine, just write it on the graph in it's relative position. For instance, -2-3^(1/2) is a little less than -2. So you just pick some arbitrary point between -2 and -3 on the x-axis and label it -2-3^(1/2), and be done with it. Forgive me if this sounds condescending, but I hope you realize 3^(1/2) is squareroot of 3.
    So it can't go through the line of origin. I don't think this is possible for me to do without a graphing calculator. I don't understand how your getting the x axis still. Your method is diffirent from my book. -2-3^(1/2) = -3.7? A little less then -2? It's a smaller number yeah..but it's closer to negative 4. I was taught to graph the parabola once you find your vertex simply plug in 0 for x and go from their. Maybe thats why I got confused on my test because when graphing everything looked strange.
    Also another problem y = 24(x + 5.5)^2 . a = 24 h = 5.5 k = 1 ?
    So to stat this graph I'd need to find the vertex is it (-5.5, 1) and y intercept is = ??? . Do I need to use that distrubitive property for this once? 0 = 24+ x^2 + 30.25 ??
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  2. #17
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    Quote Originally Posted by D@nny View Post
    So it can't go through the line of origin. I don't think this is possible for me to do without a graphing calculator. I don't understand how your getting the x axis still. Your method is diffirent from my book. -2-3^(1/2) = -3.7? A little less then -2? It's a smaller number yeah..but it's closer to negative 4. I was taught to graph the parabola once you find your vertex simply plug in 0 for x and go from their. Maybe thats why I got confused on my test because when graphing everything looked strange.
    Also another problem y = 24(x + 5.5)^2 . a = 24 h = 5.5 k = 1 ?
    So to stat this graph I'd need to find the vertex is it (-5.5, 1) and y intercept is = ??? . Do I need to use that distrubitive property for this once? 0 = 24+ x^2 + 30.25 ??
    It is possible, you just use the quadratic formula. If you can't factor a polynomial of the form ax^2 + bx + c = 0 by grouping, you plug it into the formula x = [-b +or- squareroot(b^2 - 4ac)]/2a. That would give the exact value, no graphing calculator required. Yes, -2-3^(1/2) is a little less than -2. Remember, 3^(1/2) is the same as saying the squareroot of 3, which is about 1.7. So -2-1.7=-3.7, almost 4. You also need to plug in 0 for y, to see where it cuts the x-axis.

    In the second problem, where did you get k=1? There is no k in the equation, k=0 there, unless you mistyped the equation. And yes, you will need the distributive property. 24(x+5.5)^2 = 24(x+5.5)(x+5.5) = 24(x^2 + 11x + 30.25). Then plug in x=0 for the y-intercept, and y=0 for the x-intercept. According to the original formula, the min value is (-5.5,0). Then just plot the graph with those values and you should be fine.
    Last edited by Jhevon; February 20th 2007 at 08:23 PM.
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  3. #18
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    Quote Originally Posted by Jhevon View Post
    It is possible, you just use the quadratic formula. If you can't factor a polynomial of the form ax^2 + bx + c = 0 by grouping, you plug it into the formula x = [-b +or- squareroot(b^2 - 4ac)]/2a. That would give the exact value, no graphing calculator required. Yes, -2-3^(1/2) is a little less than -2. Remember, 3^(1/2) is the same as saying the squareroot of 3, which is about 1.7. So -2-1.7=-3.7, almost 4. You also need to plug in 0 for y, to see where it cuts the x-axis.

    In the second problem, where did you get k=1? There is no k in the equation, k=0 there, unless you mistyped the equation. And yes, you will need the distributive property. 24(x+5.5)^2 = 24(x+5.5)(x+5.5) = 24(x^2 + 11x + 30.25). Then plug in x=0 for the y-intercept, and y=0 for the x-intercept. According to the original formula, the min value is (-5.5,0). Then just plot the graph with those values and you should be fine.
    We proceed by the quadratic formula, using a=1, b=4, c=1
    We get, x = [-4 +- squareroot(4^2 - 4(1)(1))]/2(1)
    simplifying we get x= -2-3^(1/2) and -2+3^(1/2) --> the x-intercepts.

    Now the y-intercept(s), we set x=0.
    So y=(0+2)^2 -3 = 4-3=1

    This is all we need for a parabola. So we end up with:


    [IMG]file:///C:/DOCUME%7E1/Jhevon/LOCALS%7E1/Temp/moz-screenshot.jpg[/IMG][/QUOTE]

    Finally I think I got it, correct me if i'm missing a step.
    So (x+2)^2 becomes (x+2)(x+2) dedistributed ? Then redistributed into x^2 + 4x + 4 - 3
    x^2 +4x +1 = 0
    A = x or 1 B = 4 C = 1.
    x = - b / 2 a but instead of what I was doing - 4 / 2 (1) I'd replace the variables with squareroot(4^2 - 4(1)(1)]/2(1) = 1.73 = X .
    I understand that part fully now . Its plugging into the formula .
    The formula your using is similar to factoring perfect square trinomials a ^ 2 + 2ab + b^2 = (a + b )^2 . That formula isn't in our book. Why is it then - 2 + - 1.7 .
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  4. #19
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    Quote Originally Posted by D@nny View Post
    We proceed by the quadratic formula, using a=1, b=4, c=1
    We get, x = [-4 +- squareroot(4^2 - 4(1)(1))]/2(1)
    simplifying we get x= -2-3^(1/2) and -2+3^(1/2) --> the x-intercepts.

    Now the y-intercept(s), we set x=0.
    So y=(0+2)^2 -3 = 4-3=1

    This is all we need for a parabola. So we end up with:


    [IMG]file:///C:/DOCUME%7E1/Jhevon/LOCALS%7E1/Temp/moz-screenshot.jpg[/IMG]
    Finally I think I got it, correct me if i'm missing a step.
    So (x+2)^2 becomes (x+2)(x+2) dedistributed ? Then redistributed into x^2 + 4x + 4 - 3
    x^2 +4x +1 = 0
    A = x or 1 B = 4 C = 1.
    x = - b / 2 a but instead of what I was doing - 4 / 2 (1) I'd replace the variables with squareroot(4^2 - 4(1)(1)]/2(1) = 1.73 = X . Why is it B^2 - 4 * a * c / 2 a . ?
    I thought I had it but got lost again with the formula.[/quote]

    It's not "dedistributed," we "expanded." Distribute is the property that allows us to multiply out. Anyway, the quadratic formula is just a formula to memorize. And it's everything divided by 2a. so it's [-b +- sqrt(b^2 - 4ac)]/2a, so the whole -b +- sqrt(b^2 -4ac) is divided by 2a. But otherwise you were right so far. Remember, it's a +-, meaning you add for one answer, and subtract for the other answer. So one value for x would be [-b + sqrt(b^2 - 4ac)]/2a and the other [-b - sqrt(b^2 - 4ac)]/2a
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  5. #20
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    a is the coefficient of x^2, b is the coefficient of x and c is the constant.
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  6. #21
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    okay...but all this is not needed the simple b / 2a to get -2 is acceptable..I'm pretty sure about that. Your formula is longer and that quadratic formula is no where in my book. Standard form of the quadratic equation is ax^2 bx + c = 0.
    I'v finally covered 7/8 of the chapter. Can you tell me why I'm not getting this plus negative thing. It wants me to find both values of x. X^2 + 6x + 9 = 1. Naturally I factor x^2 and 6 x into
    x^2 + 6x + 9 = 1.
    (x+3)^2 =1 . The square root of 1 is one.
    x + 3 = 1
    x = -4 . or x = ? The book says the 2 answers are x = -4,-2 . What's wrong with my math?
    Ps I don't solve by completing the square right? (B/2)^2 Like you did up their.
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  7. #22
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    Quote Originally Posted by D@nny View Post
    okay...but all this is not needed the simple b / 2a to get -2 is acceptable..I'm pretty sure about that. Your formula is longer and that quadratic formula is no where in my book. Standard form of the quadratic equation is ax^2 bx + c = 0.
    I'v finally covered 7/8 of the chapter. Can you tell me why I'm not getting this plus negative thing. It wants me to find both values of x. X^2 + 6x + 9 = 1. Naturally I factor x^2 and 6 x into
    x^2 + 6x + 9 = 1.
    (x+3)^2 =1 . The square root of 1 is one.
    x + 3 = 1
    x = -4 . or x = ? The book says the 2 answers are x = -4,-2 . What's wrong with my math?
    Ps I don't solve by completing the square right? (B/2)^2 Like you did up their.
    Yes, the -b/2a formula works as well, but that's only for the vertex. The longer formula is to solve for x when y is 0. It's not always necessary, just when you can't factorize by grouping. And it's next to impossible for that formula not to be in your book, look again. As for your problem, completing the square is not the way i would go. It can work, and i can show you if you want, but i'd foil. Here's how to do it:

    x^2 +6x +9=1
    => x^2 +6x + 8=0.............i subtracted 1 from both sides.
    =>(x+4)(x+2)=0................if you don't get this step, tell me, i don't want to explain it if its not necessary.

    since we have two things multipying give 0, it means one or the other must be zero. so either x+4 = 0 or x+2=0. so we have x=-4 or x=-2.
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  8. #23
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    The + - thing was a part of the formula. if you're not using the quadratic formula, dont worry about it.
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  9. #24
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    doing it your way, you have to realize that the squareroot of 1 is + or -1. And that is true for any number. squareroot of 9 is + or -3. and so forth. since a negative*negative gives a positive. so (-1)(-1)=1 just as (1)(1)=1. so the squareroot of 1 is + - 1. So you would end up with x+3=1 and x+3 = -1. so x=-2, x=-4
    Last edited by Jhevon; February 20th 2007 at 10:11 PM.
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  10. #25
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    Quote Originally Posted by Jhevon View Post
    doing it your way, you have to realize that the squareroot of 1 is + or -1. And that is true for any number. squareroot of 9 is + or -3. and so forth. since a negative*negative gives a positive. so (-1)(-1)=1 just as (1)(1)=1. so the squareroot of 1 is + - 1. So you would end up with x+3=1 and x+3 = -1. so x=-2, x=-4
    dOh! Stupid mistake. Thanks again man. That other way is even easier though. I'm going to stick to that.
    Here is a problem. X^2 + 6X + 41 = 0
    x^2 - 6x = - 41
    x^2 - 6x +9 = -41 + 9
    (x-3)^2 = -32
    x-3 = +- squareroot -32
    x = 3 +- part I'm confused on. I have to simplify the -32 squareroot by using i = -1. Right? So I need to factor which Im doing without much knowledge of. I factored - 32 but I'm not sure if I chose the right numbers. Is it x = 3 +- 2i* squareroot2 or x = 3 +- 4i*squareroot2. How do you know what numbers simplify down from -32 squarerooted? Is their a good method to do it so I won't switch up what number is inside the squareroot side with the one out? I think thats the mistake I made on the last test.
    Last edited by D@nny; February 20th 2007 at 10:44 PM.
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  11. #26
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    Quote Originally Posted by D@nny View Post
    dOh! Stupid mistake. Thanks again man. That other way is even easier though. I'm going to stick to that.
    Here is a problem. X^2 + 6X + 41 = 0 (6/2)^2 = 9
    x^2 - 6x = - 41
    x^2 - 6x +9 = -41 + 9
    (x-3)^2 = -32
    x-3 = +- squareroot -32
    x = 3 +- part I'm confused on. I have to simplify the -32 squareroot by using i = -1. Right? So I need to factor which Im doing without much knowledge of. I factored - 32 but I'm not sure if I chose the right numbers. Is it x = 3 +- 2i* squareroot2 or x = 3 +- 4i*squareroot2. How do you know what numbers simplify down from -32 squarerooted? Is their a good method to do it so I won't switch up what number is inside the squareroot side with the one out? I think thats the mistake I made on the last test.
    give me the question again, that 0 (6/2)^2 = 9 part makes no sense to me. just type the problem as it appears. You do complex numbers in your class?
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  12. #27
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    Quote Originally Posted by Jhevon View Post
    give me the question again, that 0 (6/2)^2 = 9 part makes no sense to me. just type the problem as it appears. You do complex numbers in your class?
    ignore that...I dont know how it got their..Yes we are...
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  13. #28
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    Quote Originally Posted by D@nny View Post
    dOh! Stupid mistake. Thanks again man. That other way is even easier though. I'm going to stick to that.
    Here is a problem. X^2 + 6X + 41 = 0
    x^2 - 6x = - 41
    x^2 - 6x +9 = -41 + 9
    (x-3)^2 = -32
    x-3 = +- squareroot -32
    x = 3 +- part I'm confused on. I have to simplify the -32 squareroot by using i = -1. Right? So I need to factor which Im doing without much knowledge of. I factored - 32 but I'm not sure if I chose the right numbers. Is it x = 3 +- 2i* squareroot2 or x = 3 +- 4i*squareroot2. How do you know what numbers simplify down from -32 squarerooted? Is their a good method to do it so I won't switch up what number is inside the squareroot side with the one out? I think thats the mistake I made on the last test.
    If the original question had a +6x it should be a -3 +- 4squareroot2*i. So your answer is right other than the sign. You know -32 breaks down further because you realize you can right it as the product of two numbers, one of which is a perfect square, namely, 16*2. But that is not something that should mess you up on a test. Unless there are specific instructions that say "simplify your answer," -3 +- i*squareroot32 is not wrong.
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  14. #29
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    hey man, it's 3:10 a.m here, i have to get up at 6:30 a.m. so i'm going to leave soon. if there is anything else u wanna ask, ask now
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  15. #30
    is up to his old tricks again! Jhevon's Avatar
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    Sorry Danny, gotta go get some sleep man. Hopefully someone else here can help you. good luck
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