You haven't specified any equation so let's make one up as an example: y=x^2-3x-18.
The x-axis is just the line where y=0. Put y=0 in the equation: x^2-3x-18=0. This factorises:
(x-6)(x+3)=0. So x=6 or x=-3 make this equation 0. That means that when y=0 (the x-axis), x=6 and x=-3. These are the points, (6,0) and (-3,0), where the curve cuts the x-axis.
The question also talks about the tangent. Unless you've done calculus you won't know how to do this. So let's assume you do know some calculus. If you don't, just skip the section below.
The tangent is just the slope of the curve at some point. The slope varies on all curves. Straight lines (linear graphs) have constant gradients. To find the gradient or slope we first find the derivative of the equation, usually written dy/dx. In the example it's 2x-3. Then we need to know at what value of x we need the tangent. Let's make x=4, then dy/dx=8-3=5. When x=4, y=4^2-3*4-18=16-12-18=-14. Therefore the tangent is at (4,-14) on the curve where the slope is 5.
The tangent line has a slope of 5 so we know the linear equation is of the type y=5x+c where we need to find c by putting in the tangent point: -14=5*4+c. So c=-14-20=-34 and the equation of the tangent line is y=5x-34. It cuts the x-axis when y=0, so 5x-34=0 and x=34/5=6.8. Therefore the tangent meets the x-axis at x=6.8 or the point (6.8,0).