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Math Help - Finding the zeros and intercepts---help / check answer?

  1. #1
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    Finding the zeros and intercepts---help / check answer?

    Hi, again I am rusty at math (I haven't done it in two years). There are no solutions in my book for my homework, so could someone double check that I have the right idea?


    1) Find the zeros of the given function f.
    f(x)=x^3-x^2-2x
    f(x)=x(x^2-x-2)
    f(x)=x(x-2)(x+1)

    So, are the zeros x=2, x=-1, and x=0?
    One more...

    2) Find the x- and y-intercepts, if any, of the graph of the given function f. Do not graph.

    f(x)=(2x-3)(x^2+8x+16)
    f(x)=2x^3+16x^2+32x-3x^2-24x+13
    f(x)=2x^3+13x^2+8x+13
    f(x)=x(2x^2+13x+21)
    f(x)=x(2x+7)(x+3)

    For the X-intercepts I got...
    (-7/2,0), (-3,0), (0,0)

    How do I find the y-intercepts then? Do I plug in zero for "x"? If I do, wouldn't the y-intercept be just (0,0)?

    Thank you so much in advanced!
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  2. #2
    Super Member 11rdc11's Avatar
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    yes plug in 0 for x to find y intercept. The y intercept is (0,13)
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  3. #3
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    Quote Originally Posted by 11rdc11 View Post
    yes plug in 0 for x to find y intercept. The y intercept is (0,13)
    Thank you so much! I highly appreciate it. Do the other questions look alright?
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  4. #4
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    Just to clarify, these two functions are not the same:
    f(x)=2x^3+13x^2+8x+13

    f(x)=x(2x+7)(x+3)

    The first has a y-intercept of (0, 13) and the second has a y-intercept of (0, 21). Did you try to find x-intercepts for the first equation at all?
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  5. #5
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    Quote Originally Posted by icemanfan View Post
    Just to clarify, these two functions are not the same:
    f(x)=2x^3+13x^2+8x+13

    f(x)=x(2x+7)(x+3)

    The first has a y-intercept of (0, 13) and the second has a y-intercept of (0, 21). Did you try to find x-intercepts for the first equation at all?

    Yes, I did (first post). I was unsure if it was correct.

    (-7/2, 0), (-3, 0), (0, 0)

    So you must always find the y-intercept of both functions, not just the first one? Thanks in advanced.
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  6. #6
    Super Member 11rdc11's Avatar
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    You made your mistake right here, this is not true

    2x^3+13x^2+8x+13 = x(2x^2+13x+21)



    The people who made the problem gave

    <br />
f(x)=(2x-3)(x^2+8x+16)<br />

    instead of distributing it out

    2x^3+13x^2+8x+13

    so you could find your x intercepts easier.

    0=2x-3

    0 = x^2+8x+16
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  7. #7
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    Ah! Thank you so much...I totally forgot about this. So the correct way would be:

    0=2x-3

    \frac 32 = x


    x^2+8x+16
    (x+4)(x+4)
    x=-4

    So the x-intercepts are ( \frac 32, 0) and (-4, 0)
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  8. #8
    Super Member 11rdc11's Avatar
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    Yep
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  9. #9
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    Yay! Thank you! I'm so glad someone can point out my silly mistakes, haha.
    The homework is going alright so far, but I'm a little uneasy that I have the correct way of doing this.

    Find an equation of the line that satisfies the given conditions:
    through the origin parallel to the line through (1,0) and (-2,6)

    So my way of thinking was: find the slope since it is the same as the parallel line, and then the origin is at (0,0), correct? Thus...

    m= \frac {6-0} {-2-1}

    m=-2

    plug it into the point-slope equation...

    y-0^2=m(x-0^2)

    therefore the equation would be...(I hope)

    y=-2x
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  10. #10
    Super Member 11rdc11's Avatar
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    y_2 - y_1 = -2(x_2 - x_1)

    y - 0 = -2(x - 1)

    Use either of your original points
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  11. #11
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    Quote Originally Posted by 11rdc11 View Post
    y_2 - y_1 = -2(x_2 - x_1)

    y - 0 = -2(x - 1)

    Use either of your original points
    So if it says "through the origin parallel to the line through (1,0) and (-2,6)"...we can use those points which are parallel to the line it's talking about? Texts are so awkwardly worded sometimes, it makes my head spin.

    So, the equation would be...
    y-1^2=-2x(x-0^2)
    y-1=-2x
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  12. #12
    Super Member 11rdc11's Avatar
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    Oops I didn't read the orgin part so your answer is correct

    y = -2x

    Check question, why are you squaring here

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  13. #13
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    Quote Originally Posted by 11rdc11 View Post
    Oops I didn't read the orgin part so your answer is correct

    y = -2x

    Check question, why are you squaring here

    Oh okay! Thanks!
    HAHA yes that's what I get for my bad sleeping habits. That's meant to be y2 and x2...not squared. Sorry, I glanced over the formula super fast. Ignore that...

    Anyway, I know you are sick of answering math problems. Maybe someone else will be able to help. I really hope I can get back into math in the next few weeks so I can figure things out on my own...

    I just need explaining on this one:
    "If C is an arbitrary real constant, an equation such as 2x - 3y = C is said to define a family of lines. Choose four different values of C and plot the corresponding lines on the same coordinate axes. What is true about hte lines that are members of this family?"

    Do I plug in numbers for C? If I do, how do I solve that?
    I was thinking to plug in numbers for x and y but I do not think it's asking that...
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  14. #14
    Super Member 11rdc11's Avatar
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    2x - 3y = C

    -3y = -2x + C

    y = \frac{2x}{3} -\frac{C}{3}

    Ok now put in whatever value for C

    The only thing that is happening is the graph is moving up or down. It is easier to see what I'm trying to explain. Try plooting it and you will see what I mean.
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  15. #15
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    Thank you so much! That makes way more sense. I will play around with that and see what I get. *must review notes...*
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