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calculusxy

  • one year ago

If the function f is defined by f(x) = x^2 + bx + c, where b and c are positive constants, which could be the graph of f?

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  1. calculusxy
    • one year ago
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    that seems like quadratic function, if i am not wrong. @Vocaloid

  2. calculusxy
    • one year ago
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    so it would be a parabola.

  3. Vocaloid
    • one year ago
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    yes

  4. calculusxy
    • one year ago
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    |dw:1443919129693:dw|

  5. calculusxy
    • one year ago
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    it's not the best graph, but would that be correct?

  6. Vocaloid
    • one year ago
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    I think so, as long as the graph doesn't pass through the origin if c is positive, then the y-intercept should be something bigger than 0

  7. calculusxy
    • one year ago
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    but it does pass through the origin

  8. Vocaloid
    • one year ago
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    hmmm then the answer is probably some other graph

  9. Vocaloid
    • one year ago
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    is there list of choices?

  10. calculusxy
    • one year ago
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    yes

  11. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle f(x)=x^2+bx+c }\) \(\large\color{black}{ \displaystyle f(x)=\left(x^2+bx\right)+c }\) \(\large\color{black}{ \displaystyle f(x)=\left(x^2+bx+\frac{b^2}{4}\right)-\frac{b^2}{4}+c }\) \(\large\color{black}{ \displaystyle f(x)=\left(x+\frac{b}{2}\right)^2+\left(c-\frac{b^2}{4}\right) }\) So, we see a shift to the left by b/2 units, and a shift vertically (whether up or down depends on whether b²/4 is greater smaller or equal to c)

  12. calculusxy
    • one year ago
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  13. SolomonZelman
    • one year ago
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    If, c>b\(^2\)/4 then the shift is up. If, c<b\(^2\)/4 then the shift is down. If, c=b\(^2\)/4 then no vertical shift.

  14. calculusxy
    • one year ago
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    @SolomonZelman Sorry but i didn't quite understand what you were saying. i am in the beginning of 8th grade ...

  15. SolomonZelman
    • one year ago
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    I just completed the square, that is all. And now you have two possible answers for your problems (again depending on which is greater c or b\(^2\)/4).

  16. SolomonZelman
    • one year ago
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    Maybe a brief guide with rules of a shift of a function will help? \(\large\color{ teal }{\large {\bbox[5pt, lightcyan ,border:2px solid white ]{ \large\text{ }\\ \begin{array}{|c|c|c|c|} \hline \texttt{Shifts} ~~~\tt from~~~ {f(x)~~~\tt to~~~g(x)}&~\tt{c~~~units~~~~} \\ \hline \\f(x)= x^2 ~~~~~\rm{\Rightarrow}~~~~ g(x)= (x \normalsize\color{red}{ -~\rm{c} })^2 &~\rm{to~~the~~right~} \\ \text{ } \\ f(x)= x^2 ~~~~~\rm{\Rightarrow}~~~~ g(x)= (x \normalsize\color{red}{ +~\rm{c} })^2&~\rm{to~~the~~left ~} \\ \text{ } \\ f(x)= x^2 ~~~~~\rm{\Rightarrow}~~~~ g(x)= x^2 \normalsize\color{red}{ +~\rm{c} } &~\rm{up~} \\ \text{ } \\ f(x)= x^2 ~~~~~\rm{\Rightarrow}~~~~ g(x)= x^2 \normalsize\color{red}{ -~\rm{c} } &~\rm{down~} \\ \\ \hline \end{array} }}}\)

  17. SolomonZelman
    • one year ago
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    This is a rule where c denotes a positive number. (Any real number greater than 0)

  18. calculusxy
    • one year ago
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    i truly mean it. i don't understand anything that u just said

  19. SolomonZelman
    • one year ago
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    sorry

  20. SolomonZelman
    • one year ago
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    I didn't know you are unfamiliar with completing the square, or perhaps my presenation of it was poor. I apologize again-:( Good luck tho (but, a side note there are 2 possible answers to your problem)

  21. Vocaloid
    • one year ago
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    well, let's review what we know so far

  22. Vocaloid
    • one year ago
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    f(x) = x^2 + bx + c, b and c are positive constants, correct? since the coefficient on x^2 is positive (1), we know that the parabola must face upwards, with me so far?

  23. calculusxy
    • one year ago
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    yes

  24. Vocaloid
    • one year ago
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    ok, now let's try plugging in x = 0 to find the y-intercept of the graph f(x) = x^2 + bx + c f(x) = 0^2 + b(0) + c = c so the y-intercept is (0,c), and since c is positive, the y-intercept should be positive

  25. Vocaloid
    • one year ago
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    so, out of the 5 answer choices, only one is an upwards parabola with a positive y-intercept, which one is it?

  26. calculusxy
    • one year ago
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    would it be e?

  27. Vocaloid
    • one year ago
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    yup! that's it

  28. calculusxy
    • one year ago
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    thanks!

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