The product of all the values of x for which the tangent lines to the graph of y= 3/3−x are parallel to the line 3x−10y + 1 = 0 is

Answer 1

y= 3/(3-x) has tangent line where the slope = y' = 3/(3-x)^2 = 3/10

to be parallel to 3x-10y+1 = 0 solve for y, then the slope = the coefficient of the x term

10y = 3x+1

y = (3/10)x + 1/10. Slope = 3/10. Any parallel line has slope = 3/10 and y intercept different from 1/10

3/(3-x)^2 = 3/10

(3-x)^2 = 10

3-x = + or - sqr10

-x = -3 + or - sqr10

x = 3 + or - sqr10

x = about 6.16 or - 0.16