Verified answer

To find y', use implicit differentiation:

6x² + 3xy + 2y² + 17y - 6 = 0

12x + 3y + 3xy' + 4yy' + 17y' = 0

y'(3x + 4y + 17) + 12x + 3y = 0

y'(3x + 4y + 17) = -12x - 3y

y' = (-12x - 3y) / (3x + 4y + 17)

Then plug in the values of x and y from the point you are given to find y' at that point. That will be the slope of the tangent line, and the slope of the normal line will be the negative reciprocal. Then just use the point and the slopes to write the equations of the lines in point-slope form.

Hope that helps :)

i visit get the tangent as i don't understand what do u recommend through standard line y=mx+b to get m get f' at this factor f(x)=(x^2+3)/(2x) = x/2 + 3/2x f'(x)=a million - 3/(2x^2) f'(a million)=a million-3/2= - a million/2 so m= -a million/2 y= -a million/2 x +b to get b use the factor (a million,2) a million = (-a million/2 )*2 +b a million=-a million+b b=2 so the line equation of the tangent is y= -x/2 +2