Answer:
Part a) The approximate years were 1998 and 2008
Part b) The graph in the attached figure
Step-by-step explanation:
Part a) we have
[tex]y=-0.27x^{2}+3.3x+77[/tex]
For [tex]y=80\ billion\ eggs[/tex]
Solve the quadratic equation
[tex]80=-0.27x^{2}+3.3x+77[/tex]
[tex]0.27x^{2}-3.3x+3=0[/tex]
we know that
The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]0.27x^{2}-3.3x+3=0[/tex]
so
[tex]a=0.27\\b=-3.3\\c=3[/tex]
substitute in the formula
[tex]x=\frac{3.3(+/-)\sqrt{-3.3^{2}-4(0.27)(3)}} {2(0.27)}[/tex]
[tex]x=\frac{3.3(+/-)\sqrt{7.65}} {0.54}[/tex]
[tex]x=\frac{3.3(+)\sqrt{7.65}} {0.54}=11.23\ years[/tex]
[tex]x=\frac{3.3(-)\sqrt{7.65}} {0.54}=0.99\ years[/tex]
therefore
The approximate years are
1997+11=2008
1997+1=1998
Part b)
Using a graphing tool
we have
[tex]y=-0.27x^{2}+3.3x+77[/tex]
[tex]y=80[/tex]
The solution of the system of equations is the intersection point both graphs
The intersection point are (0.99,80) and (11.23,80)
see the attached figure
therefore
The solution part a) is correct
