Find the coordinates of the points where the graph of y = 2x^3 + 3x^2 - 4x + 1 cuts the x-axis.?

Answer 1

2x^3+3x^2-4x+1=0
If any rational root p/q p=+-1 and q = +-1 and+-2
x=1/2 is a root so
y=(x-1/2) (x^2+4x-2)
the second factor has roots
x=((-4+-sqrt(24))/4 =-1+-1/2sqrt(6)

Answer 2

Hai Plod,

Any graph will cut the x axis at a point where its y co-ordinate is 0(zero) and vice versa(i. e., graph cuts y axis when x = 0)

Therefore, the curve y = 2x^3 + 3x^2 - 4x + 1, will cut the x axis if y = 0.
i. e. y = 2x^3 + 3x^2 - 4x + 1 = 0

So, the x co-ordinate of the required point is obtained by solving the equation 2x^3 + 3x^2 - 4x + 1 = 0 for x.

Answer 3

The first thing you need to do is factor.

find (2x-1)(x^2+2x-1)

You will need the quadratic formula to find the x values for the second part
[-b +- sqrt(b^2-4ac)]/2a

find the value(s) for x which causes the value in the parentheses to be = 0

That is where the graph will intercept the x-axis.

Answer 4

y=2x^3+3x^2-4=
2x3+3x2-4x= -1
x[2x^2+3x-4]= -1
x[2x+2] [x-2]= -1

x=-1
or 2x+2= -1
2x=-3
x=-3/2
or x-2= -1
x=1


i gusse that :D

Answer 5

let x=0