Explanation:
PART 1.
A quadratic function is given by the form:
[tex]f(x)=ax^2+bx+c[/tex]
While a linear function is given by:
[tex]y=mx+b[/tex]
In this case, we have 2 cases:
CASE 1
[tex]f(x)=-x(2x+8)[/tex]
Applying distributive property we get:
[tex]f(x)=-x(2x+8) \\ \\ f(x)=-2x^2-8x[/tex]
So this is a quadratic function.
[tex]\text{Quadratic term:}-2x^2 \\ \\ \text{linear term:}-8x \\ \\ \text{constant term:} None[/tex]
CASE 2
[tex]f(x)=x^2-7[/tex]
So this is a quadratic function.
[tex]\text{Quadratic term:} \ x^2 \\ \\ \text{linear term:} \ None \\ \\ \text{constant term:} \ -7[/tex]
PART 2.
Remember that for a quadratic function we have:
[tex]f(x)=ax^2+bx+c \\ \\ \\ a,b,c \ are \ constants[/tex]
From the table:
[tex]When \ x=0, \ y=9: \\ \\ f(0)=9=a(0)^2+b(0)+c \\ \\ c=9[/tex]
[tex]When \ x=1, \ y=16: \\ \\ f(1)=16=a(1)^2+b(1)+9 \\ \\ 16=a+b+9 \\ \\ a+b=7[/tex]
[tex]When \ x=-4, \ y=1: \\ \\ f(-4)=1=a(-4)^2+b(-4)+9 \\ \\ 1=16a-4b+9 \\ \\ 16a-4b=-8[/tex]
Solving for a and b:
[tex]a=7-b \\ \\ \text{Substituting into 16a-4b=-8:} \\ \\ 16(7-b)-4b=-8 \\ \\ 112-16b-4b=-8 \\ \\ -20b=-8-112 \\ \\-20b=-120 \\ \\ b=6 \\ \\ Then: \\ \\ a=7-6 \\ \\ a=1[/tex]
So the equation is:
[tex]\boxed{f(x)=x^2+6x+9}[/tex]
