Introduction:
To find the equation of the tangent to the curve y=x²-3x at the point whose abscissa is 1, we need to first find the slope of the tangent at that point. We can then use the point-slope form of the equation of a line to find the equation of the tangent.

Step 1: Find the slope of the tangent:
To find the slope of the tangent at the point (1, -2), we need to find the derivative of the function y=x²-3x and evaluate it at x=1.

y=x²-3x
dy/dx=2x-3
dy/dx=2(1)-3=-1

So the slope of the tangent at the point (1, -2) is -1.

Step 2: Use the point-slope form:
Now that we know the slope of the tangent, we can use the point-slope form of the equation of a line to find the equation of the tangent. The point-slope form is given by:

y-y1=m(x-x1)

Where m is the slope of the line and (x1, y1) is a point on the line. In this case, we can use the point (1, -2) as our point on the line, and the slope we found (-1) as the slope of the line. Plugging these values into the point-slope form, we get:

y-(-2)=-1(x-1)

Simplifying this equation, we get:

y=-x-1

So the equation of the tangent to the curve y=x²-3x at the point whose abscissa is 1 is y=-x-1.

Conclusion:
To summarize, we found the equation of the tangent to the curve y=x²-3x at the point whose abscissa is 1 by first finding the slope of the tangent using the derivative of the function, and then using the point-slope form of the equation of a line to find the equation of the tangent. The final equation we obtained was y=-x-1.