Given:
The rational function is:
[tex]f(x)=\dfrac{5-x}{x^2+5x+6}[/tex]
To find:
The points on the graph at the function value [tex]f(x)=3[/tex].
Solution:
We have,
[tex]f(x)=\dfrac{5-x}{x^2+5x+6}[/tex]
Substituting [tex]f(x)=3[/tex], we get
[tex]3=\dfrac{5-x}{x^2+5x+6}[/tex]
[tex]3(x^2+5x+6)=5-x[/tex]
[tex]3x^2+15x+18=5-x[/tex]
Moving all the terms on one side, we get
[tex]3x^2+15x+18-5+x=0[/tex]
[tex]3x^2+16x+13=0[/tex]
Splitting the middle term, we get
[tex]3x^2+3x+13x+13=0[/tex]
[tex]3x(x+1)+13(x+1)=0[/tex]
[tex](3x+13)(x+1)=0[/tex]
Using zero product property, we get
[tex](3x+13)=0\text{ or }(x+1)=0[/tex]
[tex]x=-\dfrac{13}{3}\text{ or }x=-1[/tex]
Therefore, the required values are [tex]-\dfrac{13}{3},-1[/tex].
