Question

What is the graph of the function f(x) =-x^2+x+20/x+4

Answer 1

Start by factoring the top and bottom.

Answer 2

how do i do that?

Answer 3

it is a line.

Factor the top.

Answer 4

Let me give you an example : x^2 +6x +4

This factors to (x+4)(x+2)

Answer 5

x+4 cancel on top and bottom

Answer 6

\(\bf -x^2+x+20\implies -(x^2-x-20)\)

Answer 7

thus factor the numerator.... and \(\bf f(x)=\cfrac{(\qquad )\cancel{ (\qquad ) }}{\cancel{ x+4 }}\)

Answer 8

-(+4)(x-5)?

Answer 9

-[ (x+4)(x-5) ] is correct so \(\bf f(x)=\cfrac{-[\cancel{ ( x+4) }(x-5 )]}{\cancel{ x+4 }}\)

Answer 10

so that'd be the graph of that
is a line, pick a couple of random "x" and get a "y"
plot the line :)

keeping in mind the "restrictions" of the original rational

that when the denominator is 0, the fraction becomes "undefined" and thus has a hole there due to discontinuity \(\bf f(x)=\cfrac{-x^2+x+20}{{\color{brown}{ x+4}}}\qquad {\color{brown}{ x+4}}=0\qquad \cfrac{-x^2+x+20}{{\color{brown}{ 0}}}\to discontinuous\)

Answer 11

hmm actually..... it may not be discontinuous

Answer 12

\(\huge\color{ blue }{\huge {\bbox[5pt, cyan ,border:2px solid white ]{ \LARGE\text{Chart    for    f(x) .}\\
\begin{array}{|c|c|c|c|}
\hline~~~~~~\textbf{x}~~~~~~ ~&~~~~~~\textbf{f(x) }~~~~~~\\
\hline\text{0} &\text{4} \\
\text{1} &\text{5} \\
\text{2} &\text{6} \\
\text{3} &\text{7} \\
\text{4} &\text{8} \\
\text{••••} &\text{••••} \\
\hline
\end{array} }}}\)

Answer 13

well, jdoe, it is just f(x)=x+4 , technically.

Answer 14

yeah.... when the denominator zeros out, so does the numerator....so it wouldn't be discontinous

Answer 15

none of the graphs that are answers are just like a straight line... im confused