72.
sketching the graph of a polynomial function
g(x)=-x^2+10x-16
a) applying the leading coefficient test
Consider the given function.
g(x)=-x^2+10x-16
Because the leading coefficient is negative and the degree is even, the graph eventually falls to the left and to the right.
b) finding the real zeros of the polynomial
g(x)=0
-x^2+10x-16=0............factor
-(x^2-10x+16)=0
-(x^2-2x-8x+16)=0
-((x^2-2x)-(8x-16))=0
-(x(x-2)-8(x-2))=0
-(x - 8) (x - 2) = 0
zeros: x=8 and x=2
c) plotting sufficient solution points
we need at least 4-5 points
we have x-intercepts and using given function we will find two more points
g(x)=-x^2+10x-16 ..........let x=0, then
g(x)=-16
g(x)=-x^2+10x-16 ..........let x=1, then
g(x)=-1^2+10*1-16=-7
g(x)=-x^2+10x-16 ..........let x=10, then
g(x)=-(10)^2+10*(10)-16=-16
so, the points are:
(0,-16)
(8,0)
(2,0)
(1,-7)
(10,-16)
d) and drawing continuous curve through the points
