Sketching Graphs of Polynomial Functions

Section 2.2
Question 72

Can you do 72 (a - d) as a guide for me to do a few others?

Thank you.

 

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

 

You are simply amazing. Beautiful math work. I can now use this reply to answer a few more on my own when time allows.