Calculus Derivative question?

Question

Find the equation of the tangent to y=x^2-4x+1 that is perpendicular to the line with equation 3x+ 6y-2=0

Answer 1

[Note: C-Wryte above made an almost-empty post and then merely edited to plagarize my answer. Lame!]

The slope-intercept form of the line is:

y = -1/2x + 1/3

So its slope is -1/2. A perpendicular line has slope that's the negative reciprocal of that: -1/(-1/2) = 2.

Now you want to find the tangent to the curve where the slope is 2.

The derivative which gives the slope of the tangent is:

dy/dx = d(x^2 - 4x + 1)/dx
dy/dx = 2x - 4

... and you want to find where dy/dx=2:

dy/dx = 2x - 4
2 = 2x - 4
6 = 2x
x = 3

When x=3, the original equation's y-value is:

y = x^2 - 4x + 1
y = 3^2 - 4*3 + 1
y = -2

So, you want to find the equation for a line passing through (3,-2) with a slope of 2.

y = mx + b
-2 = 2*3 + b
b = -8

The equation is:

y = 2x -8

Answer 2

Remember that the derivative of a function gives the slope of the tangent lines. Perpendicular lines have slopes that are negative reciprocals of each other. So find the slope of 3x + 6y - 2 = 0, and call it m. Then take the derivative of y = x^2 - 4x + 1 and set it equal to -1/m. Solve this for x. Use this x value to write the equation for the tangent line (use it to derive -1/m and the point (x,y)).

Answer 3

Firstly, what is the slope of the line that is perpendicular to 3x + 6y = 2? In y = mx + b form:

6y = -3x + 2, or y = -0.5x + (1/3), so the slope of this line is -1/2 or -0.5. A perpendicular line has a slope that is the negative reciprocal of this, or 2.

So there must be a line tangent to y = x^2 - 4x + 1 with a slope of 2. Let's find it.

y' = 2x - 4. The derivative gives the slope. Where is the slope = 2?

2 = 2x - 4, or x = 3. Plug x = 3 into the function to find the y-value of this point. y = (3^2) - 4(3) + 1 = -2. So the point is (3, -2).

We have a point and the slope of the line, so we can find its equation:

y- y1 = m(x - x1)
y + 2 = 2(x - 3)
y = 2x - 8

Answer 4

The tangent at x has gradient 2x - 4 .......(1)
Rearranging 3x + 6y - 2 = 0,
y = (1/3) - (1/2)x
The gradient is -1/2
If this is to be perpendicular to the tangent, then the tangent must have gradient 2.
From (1):
2x - 4 = 2
2x = 6
x = 3.
The tangent is therefore at the point (3, 3^2 - 12 + 1) = (3,-2).
and its equation is:
y = 2x + k .......(2)
where k must be chosen so that y = -2 when x = 3.
-2 = 6 + k
k = -8.
Putting this in (2) gives the equation:
y = 2x - 8.

Answer 5

OK you just have to think through this logically.

First of all, you know the slope is equal to the derivative of the function, so we need to find the derivative, which is simple:

y = x^2 - 4x + 1
y' = 2x - 4

Now, we need the slope of the line, and then we need the opposite reciprocal for the perpendicular slope:

3x + 6y - 2 = 0
6y = -3x + 2
y = -(1/2)x + 1/3

So, the slope of that line is -1/2, so the slope of a perpendicular line is equal to 2.

So, we set this equal to the derivative and find the corresponding x-value:

2 = 2x - 4
6 = 2x
x = 3.

So, the line must hit the graph at x = 3. So now, we need to find what y-value this would be at by plugging into the function:

y = x^2 - 4x + 1
y = 3^2 - 4(3) + 1
y = 9 - 12 + 1
y = -2

So, the tangent line would intersect the curve at (3,-2). Now, we just need to plug these in with the slope of 2 we found to get the equation of the line:

y = mx + b
-2 = 2*3 + b
-2 = 6 + b
b = -8

So, we have our slope and b-value, so in slope-intercept form, we get a final answer of

y = 2x - 8
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