Tangency to a curve

syndicate · #1 $Old$ November 3rd, 2009, 07:32 AM

The equation of a line tangent to the curve y^2 - 2x - 4y =1 at (-2,1) is what?

The tangent line to the curve 3x^4 - 10x + 3 at x = 1 interets the x-axis at what point?

VonNemo19 · #2 $Old$ November 3rd, 2009, 07:53 AM

Quote:

Originally Posted by syndicate $View Post$

The equation of a line tangent to the curve y^2 - 2x - 4y =1 at (-2,1) is what?

The tangent line to the curve 3x^4 - 10x + 3 at x = 1 interets the x-axis at what point?

You must impicitly differentiate in te first problem, solve for $\frac{dy}{dx}$ and evaluate at the given pointto find the slope of the tangent line. Then, using the given point along with the slope, develope the equation for the line.

For the second, evaluate $\frac{dy}{dx}$ at $x=1$ to find the slope of the tangent line. then go back and plug 1 into the original function to have the necessary point needed to develop the equation to the tangent line. Then find the x-intercept of this line.

syndicate · #3 $Old$ November 3rd, 2009, 09:47 AM

So for the first problem
dy/dx [2y-4] - 2 = 0
dy/dx = 2/2y-4 = 2/2(1) -4
dy/dx = 2/-2 = 1 = slope

y = mx + b
1 = (-1)(-2) + b
1 = 2 + b
-1 = b

y = -x -1 ?

Second problem:
12x^3 - 10 = dy/dx
dy/dx = 12(1)^3 - 10
dy/dx = 12 - 10 = 2 is the slope of the tangent line

3(1)^4 - 10 + 3
3 - 10 + 3 = -10 = y

y = mx + b
-10 = 2(1) + b
b = -12

y = 2x -12 is the slope of the tangent line?