Find the equations of the tangent and normal to the graph of the given function at the given point.

1. y+3x^2 - 2x + 1, (2,9)
2. y=1=3 square root of x, (4,7)
3.y = x square root of x - 1 , (5,10)
4. y^2 = x^3 over 4-x , (2,2)
5. y = 2 over x , (1,2)

Question

Find the equations of the tangent and normal to the graph of the given function at the given point. 

1. y+3x^2 - 2x + 1, (2,9)
2. y=1=3 square root of x, (4,7)
3.y = x square root of x - 1 , (5,10)
4. y^2 = x^3 over 4-x , (2,2)
5. y = 2 over x , (1,2)

larseighner · Answer

Find the slope of the tangent line by taking the derivate.  The slope of the normal line is -1/m when m is the slope of the tangent line.  Then just plug into the point-slope formula.

larseighner · Answer

Have you mistyped the first one? I don't see an = sign.

larseighner · Answer

And there seems to be one too many = signs in the second.

anonymous · Answer

mistyped both. sorry. im doing what you told me:) 

y=3x^2 -2x +1 (2,9)

and 

y=1+ 3 square root of x, (4,7)

larseighner · Answer

Then they are all solved for y, you can differential to find dy/dx, which is the slope of the tangent.  The slope of the normal is the negative reciprocal of that

larseighner · Answer

What do you get for the derivative of the first one?