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Math Help - horizontal or vertical?

  1. #1
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    horizontal or vertical?

    The graph of y=5X^2-5 is the graph of y=X^2-1 stretched vertically or horizontally?


    I get that it streches by 5 but how do i tell if its vertical or horizontial?

    also

    What are the x and y intercepts of F(x)=4xsquared-1?


    i got +or-1/8 for the x intercept but im sure its wrong. I thought i got it in class but i guess not. Why is math pure 30 required for college
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  2. #2
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    Quote Originally Posted by pepperroni View Post
    The graph of y=5X^2-5 is the graph of y=X^2-1 stretched vertically or horizontally?


    I get that it streches by 5 but how do i tell if its vertical or horizontial?

    also
    ....

    1) Exclude the constants.
    2) Plot/sketch/graph both functions
    3) What would you have to do to get y=x^2 to overlay y=5x^2?
    Would you stretch it horizontally or vertically?
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  3. #3
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    thanks i dont know why i didnt think of that... but is the only way to find if its vert or horizontial is by comparing? this is just me being curious now.
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  4. #4
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    Any thing done before the "main function" changes x (and so affects the graph horizontally), anything done after the "main function" changes y (and so affects the graph vertically).

    Here your "main function" is given as y= x^2- 1. 5x^2- 5= 5(x^2- 1) multiplies by 5 after doing x^2- 1 and so affects the graph vertically.

    The x-intercepts are where the graph "intercepts" the x-axis and so y= 0. y= 4x^2- 1= 0 is the same as 4x^2= 1 or x^2= \frac{1}{4}. The x-intercepts are at x= \frac{1}{2} and x= -\frac{1}{2}. Some people refer to the "x-intercepts" as just the x value ( \frac{1}{2} and -\frac{1}{2}) and some as the points ( \left(\frac{1}{2}, 0\right) and \left(-\frac{1}{2}, 0\right). Check with your teacher or textbook to see which convention is used in your class.

    The y-intercepts are where the graph "intercepts" the y-axis and so x= 0. Just put x= 0 into y= 4x^2- 1 to get y= -1. Again, the y-intercept is either the value y= -1 or the point (0, -1) depending on the convention used.

    I have no idea what "math pure 30" is but (liberal arts) colleges want you to be educated in a broad range of subjects. You will very likely be required to take at least one mathematics course in college.
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    Any thing done before the "main function" changes x (and so affects the graph horizontally), anything done after the "main function" changes y (and so affects the graph vertically).

    Here your "main function" is given as y= x^2- 1. 5x^2- 5= 5(x^2- 1) multiplies by 5 after doing x^2- 1 and so affects the graph vertically.

    The x-intercepts are where the graph "intercepts" the x-axis and so y= 0. y= 4x^2- 1= 0 is the same as 4x^2= 1 or x^2= \frac{1}{4}. The x-intercepts are at x= \frac{1}{2} and x= -\frac{1}{2}. Some people refer to the "x-intercepts" as just the x value ( \frac{1}{2} and -\frac{1}{2}) and some as the points ( \left(\frac{1}{2}, 0\right) and \left(-\frac{1}{2}, 0\right). Check with your teacher or textbook to see which convention is used in your class.

    The y-intercepts are where the graph "intercepts" the y-axis and so x= 0. Just put x= 0 into y= 4x^2- 1 to get y= -1. Again, the y-intercept is either the value y= -1 or the point (0, -1) depending on the convention used.

    I have no idea what "math pure 30" is but (liberal arts) colleges want you to be educated in a broad range of subjects. You will very likely be required to take at least one mathematics course in college.
    How did you go from 1/4 and get 1/2 and -1/2 thats were im lost i get the idea and i got that far myslef but then im not sure how to get 1/4 to the two answers you got..... sorry to be a bother... im horrible at math, but I have to make it through this class. Pure 30 is our high schools hardest math class... our teacher says the average for the class is 60 and many will not pass XD
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  6. #6
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    Quote Originally Posted by pepperroni View Post
    How did you go from 1/4 and get 1/2 and -1/2 thats were im lost i get the idea and i got that far myslef but then im not sure how to get 1/4 to the two answers you got..... sorry to be a bother... im horrible at math, but I have to make it through this class. Pure 30 is our high schools hardest math class... our teacher says the average for the class is 60 and many will not pass XD
    4x^2 - 1 = 0

    (2x - 1)(2x + 1) = 0

    set each factor equal to 0 and solve for x.
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