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Ploting points, tangent lines

  • Thread starter bonzy87
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  • #1
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Verify that (1,1) is a point on the graph of y + ln xy = 1 and find the equation of the tangent line at (1,1) to this graph

how do you go about answering this?
 

Answers and Replies

  • #2
Defennder
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Well can you do the first part, where you have to verify? What does it mean for a point (x0,y0) to be on the graph of f(x,y)?

For the second, you would need the slope of the tangent line. Now what must be done to obtain the slope of a graph of f(x) at the point say, x=x0 ? This can be done analytically without plotting the graph.
 
  • #3
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dont no how to do either parts very new to this stuff
 
  • #4
Defennder
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Ok, then let's take an example. Consider the graph of y = f(x) = x^2. Suppose I tell you the point (1,1) lies on the graph. How would you verify that. You do that by simply showing that (1,1) satisfies the equation of the graph f(x). Let x=1, y=1 and substitute these values into y = x^2. You'll find that it satisfies the equation. On the other hand, the point (2,1) does not lies on the graph, because x=2, y=1 does not satisfy the equation.

For the 2nd part, consider a graph of y=f(x). The gradient of the graph at the point x=a would be [itex]y'(a)[/itex] The notation y' just means [tex]\frac{dy}{dx}[/tex]. Now the equation of a straight line is y=mx + c, where m and c is the gradient and the y-intercept to be determined. You can find the gradient of the graph f(x) at x=a by evaluating y'(a). To find c, just substitute in the coordinates of the point where you want to find the tangent line, as well as m, into the equation of the straight line and solve for c. That's how it's done.
 

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