Equation of tangent line to (6-2x)^1/2 at x=1

[Vladd29]

Trying to find and equation of a tangent line to (6-2x)^1/2 at x=1. For slope I’m getting -1/2, for x and y values I’m getting (1,2). The final equation I’m getting is not an answer my setup up was y-2=-1/2(x-1). Am I doing this right?

 

[Deleted member 4993]

Trying to find and equation of a tangent line to (6-2x)^1/2 at x=1. For slope I’m getting -1/2, for x and y values I’m getting (1,2). The final equation I’m getting is not an answer my setup up was y-2=-1/2(x-1). Am I doing this right?

y = (6-2x)^1/2

dy/dx = 1/2 *(-2) * (6-2x)^(-1/2)

at x = 1 → dy/dx = -1/2 and y = 2

equation of the tangent line:

y - 2 = (-1/2) * (x - 1) or

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

Looks good to me.....

 

[tkhunny]

Trying to find and equation of a tangent line to (6-2x)^1/2 at x=1. For slope I’m getting -1/2, for x and y values I’m getting (1,2). The final equation I’m getting is not an answer my setup up was y-2=-1/2(x-1). Am I doing this right?

...except that "(6-2x)^1/2" doesn't mean anything. Is that supposed to be a function definition or just an expression floating in space?

 

[Steven G]

Trying to find and equation of a tangent line to (6-2x)^1/2 at x=1. For slope I’m getting -1/2, for x and y values I’m getting (1,2). The final equation I’m getting is not an answer my setup up was y-2=-1/2(x-1). Am I doing this right?

First, you really need to what equals (6-2x)^1/2. It could be that y² = (6-2x)^1/2 or any number of expressions.

Just because your answer does not look like what you want it still can be right. Any line can be uniquely formed with two points. You can plug in to different x's (including x=1 since we know y=2) and find the correspond y values in both equations (your equation and the equation from your answer sheet). If you get the same y values from each equation (for both x's) then you have the correct answer, just in a different form.

Now if your answer does not look like the answer sheet's answer then you need to change the way your equation looks (this is why we ask for the EXACT question to be posted)

For example. if you want/need the answer to be in the form of y=....

y-2=-(1/2)(x-1)
y= -(1/2)(x-1) + 2
y= -(1/2)x +1/2 + 2
y= -(1/2)x +5/2 OR y = (-1x + 5)/2

 

[Vladd29]

y = (6-2x)^1/2

dy/dx = 1/2 *(-2) * (6-2x)^(-1/2)

at x = 1 → dy/dx = -1/2 and y = 2

equation of the tangent line:

y - 2 = (-1/2) * (x - 1) or

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

Looks good to me.....

i figured out my problem, I was multiplying the equation wrong. Thanks for your help!

 

[lookagain]

Trying to find and equation of a tangent line to (6-2x)^1/2 at x=1.

As others referred to, for this to make sense/work out, it must be written out equivalent to:

y = (6-2x)^(1/2)


The grouping symbols around the fractional exponent are required because of the Order of Operations.

Otherwise, whenever any of you above have typed that without the grouping symbols, that is equivalent to \(\displaystyle \ \ \dfrac{(6 - 2x)^1}{2}\).