Find x-and y-Intercepts of Each Graph

Discussion in 'Other Pre-University Math' started by nycmathguy, Aug 11, 2021.

  1. nycmathguy

    nycmathguy

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    Set 1.5
    David Cohen
    Questions 22 & 24

    Exercises 21–24, each figure shows the graph of an equation. Find the x- and y-intercepts of the graph. If an intercept involvesa radical, give both the radical form of the answer and a calcu-
    lator approximation rounded to two decimal places. (Check to see that your answer is consistent with the given figure.)

    Let me see.

    To find the x-intercept, let y = 0 and solve for x.
    To find the y-intercept, let x = 0 and solve for y.

    Yes?

    20210810_201246.jpg
     
    nycmathguy, Aug 11, 2021
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  2. nycmathguy

    MathLover1

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    yes
     
    MathLover1, Aug 11, 2021
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  3. nycmathguy

    nycmathguy

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    Not so bad. I will show my work when time allows.
     
    nycmathguy, Aug 11, 2021
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  4. nycmathguy

    nycmathguy

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    [QUOTE="MathLover1,


    Question 22

    y = x^3 + x^2 - 3x

    Let x = 0

    y = (0)^3 + (0)^2 - 3(0)

    y = 0.

    Our y-intercept is found at the point (0, 0) aka the origin. The graph shows this to be true.

    Let y = 0

    0 = x^3 + x^2 - 3x

    0 = x(x^2 + x - 3)

    Set each factor to 0 and solve for x.

    I see that x = 0 is one of the x-intercept found at the point (0,0). This graph shows this to be true.

    x^2 + x - 3 = 0

    By the quadratic formula, I found two more x-intercepts.

    x = -(1/2) - sqrt13}/2

    x = sqrt{13}/2 - (1/2)

    The graph shows this to be true as well.
     
    nycmathguy, Aug 11, 2021
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  5. nycmathguy

    nycmathguy

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    Question 24

    x^3 + x^2 y = 12

    Let x = 0

    (0)^3 + (0)^2 y = 12

    0 + 0 does not equal 12. I conclude there are no y-intercepts. The graph shows this to be true.

    Let y = 0

    x^3 + x^2 (0) = 12

    x^3 = 12

    Let cr = cube root

    cr{x^3} = cr{12}

    x = 2.2894284851

    Rounding to two decimal places, I get 2.29.
    The graph shows this to be true as the graph crosses the x-axis slightly to the right of 2.

    You say?
     
    nycmathguy, Aug 11, 2021
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  6. nycmathguy

    MathLover1

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    correct
     
    MathLover1, Aug 11, 2021
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  7. nycmathguy

    nycmathguy

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    Chapter 1 has been very decent so far. The David Cohen textbook is a bit more advanced and challenging. As I said many times, I am using both textbooks (Larson & Cohen) for different reasons. Larson is the book I am using to complete our precalculus trip. The Cohen book is more in terms of exploring challenging problems. As you will see, most of the Cohen textbook problems are probably meant for an honors precalculus course in high school or for a college level math course in intermediate algebra. This is why I do not weekly post from the Cohen book.
     
    nycmathguy, Aug 11, 2021
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