To find the x-intercept, substitute in 0 for y and solve for x.

`0=x^3+2x^2-13x+10`

Since x is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation.

`x^3+2*x^2-13*x+10=0`

Multiply all terms in the polynomial `x^3+2*x^2-13*x+10` `x^3+2x^2-13x+10`

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To find the x-intercept, substitute in 0 for y and solve for x.

`0=x^3+2x^2-13x+10`

Since x is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation.

`x^3+2*x^2-13*x+10=0`

Multiply all terms in the polynomial `x^3+2*x^2-13*x+10`

`x^3+2x^2-13x+10`

Use the quadratic formula to find the solutions to the equation.

`x=1,-5,2 `

The y-coordinate of any point on the x-axis is 0. The curve y = x^3+2*x^2-13*x+10 can intersect the x-axis 3 times as the highest power of x is 3. But it could be lower if the equation x^3+2*x^2-13*x+10= 0 has complex roots.

Solving the equation x^3+2*x^2-13*x+10 = 0, gives:

x^3+2*x^2-13*x+10 = 0

=> x^3 - x^2 + 3x^2 - 3x - 10x + 10 = 0

=> x^2(x - 1) + 3x(x - 1) - 10(x - 1) = 0

=> (x - 1)(x^2 + 3x - 10) = 0

=> (x -1)(x^2 + 5x - 2x - 10) = 0

=> (x -1)(x(x + 5) - 2(x + 5)) = 0

=> (x -1)(x - 2)(x + 5) = 0

=> x = 1, x = 2 and x = -5

**The curve y = x^3+2*x^2-13*x+10 has three x-intercepts at x = 1, x = 2 and x = -5**