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?

- calculusxy

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- calculusxy

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

- calculusxy

so it would be a parabola.

- Vocaloid

yes

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## More answers

- calculusxy

|dw:1443919129693:dw|

- calculusxy

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

- 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

- calculusxy

but it does pass through the origin

- Vocaloid

hmmm then the answer is probably some other graph

- Vocaloid

is there list of choices?

- calculusxy

yes

- 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)

- calculusxy

##### 1 Attachment

- SolomonZelman

If, c>b\(^2\)/4 then the shift is up.
If, c

- calculusxy

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

- 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).

- SolomonZelman

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

- SolomonZelman

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

- calculusxy

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

- SolomonZelman

sorry

- 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)

- Vocaloid

well, let's review what we know so far

- 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?

- calculusxy

yes

- 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

- Vocaloid

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

- calculusxy

would it be e?

- Vocaloid

yup! that's it

- calculusxy

thanks!

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