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.
Algebra.Com
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.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
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:
Purple Math - Synthetic Division
Make sure you read all four pages.
John

My calculator said it, I believe it, that settles it
RELATED QUESTIONS
A curve has an equation of {{{y=xe^x}}}. The curve has a stationary point at P.... (answered by Alan3354)
The curve y = e^x + 4e^-2x has one stationary point.
Find the x-coordinate of this... (answered by rothauserc)
The graph of y = 2x^3 + ax^2 + b has a stationary point (-3,19) . Find the value of a and (answered by greenestamps)
please help me to find the equation of the graph which meets the x-axis at -4 and 2, that (answered by Edwin McCravy)
The equation of the tangent line to the curve y=x^3-6x^2 at its point of inflection is
(answered by Alan3354)
a curve (dy/dx) = x^1/2-x^-1/2 the curve passes through point (4,2/3)
•find the... (answered by Fombitz)
Find the coordinates of the stationary points of the curve y = x^3(x−4) and... (answered by Fombitz)
i) Find the stationary point
ii) Find the values of x
iii) Find the maximum
(answered by ikleyn)
The equation of the tangent line to the curve y=x^3-6x^2 at its point of inflection is
(answered by jim_thompson5910)