Step-by-step explanation
A piecewise function formed by two equations with domains a≤x
In this case the two equations are:
![\begin{gathered} f(x)=2x+3n \\ h(x)=2x^2-3x-9 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/4kynxnlte6c4d4218y46ge33p4qp4gpl8g.png)
They are both polynomials so we know they are continuous in their respective domains. Then we just need to compare the values of f and h at x=-1 which is the value where their domains meet. Then the first step would be evaluating the first equation at x=-1:
![f(-1)=2\cdot(-1)+3n=-2+3n](https://img.qammunity.org/2023/formulas/mathematics/college/a3vvltzcgn0qwdoxe5tg6hnrl07jpv6l7d.png)
Then we evaluate the second equation at x=-1:
![\begin{gathered} h(-1)=2\cdot(-1)^2-3\cdot(-1)-9 \\ h(-1)=-4 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/tlj49zg3qc1ii4wsonbsph0zlm3z366kiu.png)
Then we have to look for the value of n that makes h(-1)=f(-1). Then we equalize these two results:
![-2+3n=-4](https://img.qammunity.org/2023/formulas/mathematics/college/87jeu4flcoruel164txk0t1058c0tzlcbo.png)
And we solve this equation for n. We can start by adding 2 to both sides:
![\begin{gathered} -2+3n+2=-4+2 \\ 3n=-2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/yemfgb8qmlscousjyunjg6mhj2tdqnbedc.png)
And we divide both sides by 3:
![\begin{gathered} (3n)/(3)=-(2)/(3) \\ n=-(2)/(3) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/gjep96hr19kdj19p5yiekeeqxwh2i075jw.png)
Answer
The first step to find n is evaluating the first equation for x=-1. Then the answer is the fourth option.