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Not sure if this is the right way to solve.

The question ask to sketch the graph of each function by transforming the graph of an appropriate function of the form $y=x^n$. Indicate all $x$- and $y$- intercepts on each graph.

Is this right?

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3 Answers 3

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  1. $-x^3$ is the reflection about y-axis of $x^3$
  2. $-x^3+27$ is just a transformation in positive y direction
  3. to get the x-intercept we put $y=0$ and then $$-x^3+27=0\implies -(x-3)(x^2+3x+9)=0\implies x=3$$

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  • $\begingroup$ How do I get the the x intercept for a cube? $\endgroup$
    – Prologue
    Apr 6, 2014 at 13:15
  • $\begingroup$ So my x intercepts would be x=3,x=(-/+)3 + (5)/(2)? $\endgroup$
    – Prologue
    Apr 6, 2014 at 13:21
  • $\begingroup$ x intercepts are the real solutions only $\endgroup$
    – Semsem
    Apr 6, 2014 at 13:22
  • $\begingroup$ your factorization is incorrect $\endgroup$
    – Semsem
    Apr 6, 2014 at 13:23
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    $\begingroup$ Whatever you accept my answer or not, you have to know that both upvotes of questions and answers and acceptance of answers do not decrease your reputation points. Only downvotes do. So it is nice to accept and upvote answers and questions to encourage others $\endgroup$
    – Semsem
    Apr 6, 2014 at 14:15
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To get a more accurate graph, you'll also want to find the $x$-intercept(x): where your function value $f(x) = 0$. $$27-x^3 = 0 \iff x^3 - 27 = 0 \iff (x-3)(x^2 + 3x + 9) = 0$$

This will happen at $x = 3$. ($x^2 + 3x + 9$ has no real roots, so there is one and only one real-valued x-intercept.) So you can graph the point $(3, 0)$ on your graph, as well.

*Note: your factoring is a little bit off.

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  • $\begingroup$ Yes, I notice that. Nonetheless, I see you got rid of the negative sign in front of the x by just inverting the two terms. What is that called? Or is that just your method of getting rid of the negative? $\endgroup$
    – Prologue
    Apr 6, 2014 at 13:34
  • $\begingroup$ I did that simply in the case when we set $27 - x^3$ equal to zero to find the $x$-intercept(s). $27 - x^3 = 0 \iff -(x^3 - 27) = 0 \iff x^3 - 27 = 0$. $\endgroup$
    – amWhy
    Apr 6, 2014 at 13:37
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Semsem has provided the graphs; here is how you would find the intercepts.

For the $y$ intercept, this is just when $x$ is evaluated at $0$, so plug in $x = 0$ to get $27$.

For the $x$ intercepts, factor the expression $-x^3 + 27$. Note that the factorization you provided is incorrect, you forgot to distribute the negative. The correct answer would be $-(x^3 - 27) = -(x - 3)(x^2 + 3x + 9)$. Then use zero product property and quadratic formula.

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  • $\begingroup$ Should one always take the negative sign out first when factoring? Is that a good rule? $\endgroup$
    – Prologue
    Apr 6, 2014 at 13:22
  • $\begingroup$ If you're more comfortable with having the leading coefficient be positive (which most people are), then I would suggest you do. $\endgroup$
    – MT_
    Apr 6, 2014 at 13:48

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