find the coordinates of all points where the curve y=x^3+2x^2-15x cuts the x axis
To find the coordinates of the points where the curve intersects the x-axis, we need to find the values of x when y equals zero.
Plug y=0 into the equation:
0 = x^3 + 2x^2 - 15x
Factor out an x:
x(x^2 + 2x - 15) = 0
Set each factor equal to zero:
x = 0
x^2 + 2x - 15 = 0
To solve the quadratic equation, we can factor it or use the quadratic formula: x = (-b ± √(b^2-4ac))/(2a)
In this case, factoring is simpler:
(x + 5)(x - 3) = 0
Setting each factor equal to zero, we have:
x + 5 = 0 → x = -5
x - 3 = 0 → x = 3
Therefore, the points where the curve intersects the x-axis are: (-5, 0) and (3, 0).
To find the points where the curve intersects the x-axis, we need to solve for y=0 in the equation y=x^3+2x^2-15x.
Step 1: Set y=0
0 = x^3 + 2x^2 - 15x
Step 2: Factor out common terms
0 = x(x^2 + 2x - 15)
Step 3: Factor the quadratic expression
0 = x(x + 5)(x - 3)
Step 4: Set each factor equal to zero and solve for x
x = 0, x = -5, x = 3
Therefore, the curve y = x^3 + 2x^2 - 15x intersects the x-axis at the points (0, 0), (-5, 0), and (3, 0).