Question

What are the coordinates of the points of intersection for these set of curves: `y=x(2x+5) and y= x(1+x)^2`?

Answer 1

`(-2, -2); (2, 18); (0,0)`

Explanation:

Substitute the first equation into the second.

`x(2x + 5) = x(1 + x)^2`

`2x^2 + 5x = x(1 + 2x + x^2)`

`2x^2 + 5x = x + 2x^2 + x^3`

`0 = x^3 + 2x^2 - 2x^2 + x - 5x`

`0 = x^3 - 4x`

`0 = x(x^2 - 4)`

`x = 0 and +- 2`

`:.y = 0(1 + 0)^2` and `y = 2(1 + 2)^2` and `y = -2(1 - 2)^2`

`y = 0 and 18 and -2`

Hence, the solution set is `(-2, -2), (2, 18), (0, 0)`.

Hopefully this helps!

Answer 2

The points of intersection are: `(-2,-2), (0, 0), and (2,18)`

Explanation:

Because the y coordinate of one function must equal the y coordinate of the other function (at the point of intersection), we can set the right sides of the two equations equal:

`x(2x + 5) = x(x + 1)^2`

Before we divide both sides by x, please observe that this common factor makes `x = 0` the x coordinate of one of the points of intersection: `(0, 0)`

Divide both sides by x:

`2x + 5 = (x + 1)^2`

Expand the square:

`2x + 5 = x^2 + 2x + 1`

add -2x - 1 to both sides:

`4 = x^2`

Perform the square root operation on both sides:

`x = -2 and x = 2`

check x = -2:

`y = (-2)(2(-2) + 5) = -2(1) = -2`
`y = (-2)(-2 + 1)^2 = -2(-1)^2 = -2`

The point `(-2, -2)` checks

check x = 2:

`y = 2(2(2) + 5) = 2(9) = 18`
`y = (2)(2 + 1)^2 = 2(3)^2 = 18`

The point `(2, 18)` checks.

Answer 3

(-2, -2), (0,0) and (2,18)

Explanation:

The two curves will intersect at points, obtained by solving the equation `x(2x+5)=x(1+x)^2`

`2x^2 +5x= x(1+2x+x^2)`

`2x^2 +5x = x +2x^2 +x^3`

`x^3 -4x=0`

`x(x^2-4)=0`

x(x-2)(x+2)=0`->` x= -2, 0 and 2

The corresponding y values would be -2,0 and 18

The points for intersection would thus be (-2, -2), (0,0) and (2,18)