SOLUTION: Part of the graph to a curve has equation y=x³-15/2x²+12x-18.
(a) Find the coordinates of the stationary point E.
(b) Find the coordinated of G.
G is on the X axis so Y=0.
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-> SOLUTION: Part of the graph to a curve has equation y=x³-15/2x²+12x-18.
(a) Find the coordinates of the stationary point E.
(b) Find the coordinated of G.
G is on the X axis so Y=0.
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Question 397318: Part of the graph to a curve has equation y=x³-15/2x²+12x-18.
(a) Find the coordinates of the stationary point E.
(b) Find the coordinated of G.
G is on the X axis so Y=0.
So far I have that
y=x³-15/2x²+12x-18
so dy/dx of y=x³-15/2x²+12x-18= 3²-15x+12
Now, At SP, dy/dx=0
∴ 3x²-15x+12=0
÷3
x²-5x+4=0
(x-4)(x-1)=0
x=4 or x=1
Sub x=4 into x³-15/2x²+12x-18 and x=1 into x³-15/2x²+12x-18
4³-15/2(4)²+12(4)-18. 1³-15/2(1)+12(1)-18
64-120+48-18. 1-15/2+12-18
=-26. =-25/2
(4,-26). (1,-25/2)
Now I am compltely stuck.
Can someone please help me.
It should be pretty obvious which of the two points is E. If E is the local maximum as marked on your graph, then the coordinates of E are . Otherwise the coordinates of E are
But just to make sure, take the second derivative:
Then
So is a local maximum, and
So is a local minimum.
If G is on the x-axis, then G is the single real root. If the x-coordinate of G is a rational number, then the Rational Roots Theorem says that the root must be or
The process is to use synthetic division to test each of the potential rational roots until you find one or have tested them all and concluded that there is no rational root. In this case there is one, but I'll let you find it for yourself. If you need a refresher on synthetic division, go to:
