## 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?

1. calculusxy

that seems like quadratic function, if i am not wrong. @Vocaloid

2. calculusxy

so it would be a parabola.

3. Vocaloid

yes

4. calculusxy

|dw:1443919129693:dw|

5. calculusxy

it's not the best graph, but would that be correct?

6. Vocaloid

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

but it does pass through the origin

8. Vocaloid

hmmm then the answer is probably some other graph

9. Vocaloid

is there list of choices?

10. calculusxy

yes

11. SolomonZelman

$$\large\color{black}{ \displaystyle f(x)=x^2+bx+c }$$ $$\large\color{black}{ \displaystyle f(x)=\left(x^2+bx\right)+c }$$ $$\large\color{black}{ \displaystyle f(x)=\left(x^2+bx+\frac{b^2}{4}\right)-\frac{b^2}{4}+c }$$ $$\large\color{black}{ \displaystyle f(x)=\left(x+\frac{b}{2}\right)^2+\left(c-\frac{b^2}{4}\right) }$$ 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

13. SolomonZelman

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

@SolomonZelman Sorry but i didn't quite understand what you were saying. i am in the beginning of 8th grade ...

15. SolomonZelman

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

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

This is a rule where c denotes a positive number. (Any real number greater than 0)

18. calculusxy

i truly mean it. i don't understand anything that u just said

19. SolomonZelman

sorry

20. SolomonZelman

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

well, let's review what we know so far

22. Vocaloid

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

yes

24. Vocaloid

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

so, out of the 5 answer choices, only one is an upwards parabola with a positive y-intercept, which one is it?

26. calculusxy

would it be e?

27. Vocaloid

yup! that's it

28. calculusxy

thanks!