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Few Problems

  1. Jul 1, 2008 #1
    I have multiple problems :

    Find the tangent to the curve [tex]y=x^{2}[/tex] going through the point [tex](2,0)[/tex]. Note that [tex](2,0)[/tex] is not a point of the curve.

    Well, I know the line tangent to the curve at 2 must have the slope of 2x, or 4. Now to find the equation of the line, I substitute the point into the equation y=4x + b, in which I obtain y = 4x - 8.

    I don't understand how the book obtained y= 8x -16 as the answer? What did I do wrong?

    At what point of the curve [tex]y=x^{2}[/tex] is the tangent parallel to the secant drawn through the points with abscissas 1 and 3?

    I understand the entire question until it hits to, "secant drawn through the points with abscissas..." I don't understand how I can find a single line parallel to two other lines with two different slopes ?

    For what values of b and c does the curve [tex]y=x^{2} + bx +c[/tex]have the line y=x as a tangent at the point with abscissa 2?

    No clue.
  2. jcsd
  3. Jul 1, 2008 #2


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    No, you don't know that. The line tangent to the curve at x= a must have slope 2a but the "tangent to the curve through (2, 0)" is NOT tangent to the curve AT x= 2! The problem specifically tells you that (2, 0) is NOT a point of y= x2 so the tangent uyou want is NOT tangent at x= 2.

    You don't know where that line is tangent to the curve so just call it x= a. The line tangent to y= x2 at (a, a2 is of the form y= (2a)(x- a)+ a2. For that line to pass through (2, 0), it must satisfy 0= (2a)(2-a)+ a2. Solve that equation for a.

    There are not two other lines. There is a single line given by two points. What are the points on the curve with abscissas 1 and 3? What is the slope of the line (secant) through those points?

    The point on the curve with abscissa 2 is, of course, (2, 4+ 2b+ c). What is the equation of the tangent line at that point (it will involve b and c). Set that equal to x (for all x) and solve for b and c. (Since it is true for all x, you can get two equations for a and b by letting x be any two values.)
  4. Jul 1, 2008 #3


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    Let's just take the first one. They TOLD you (2,0) was NOT on the curve y=x^2 and you went ahead and computed the slope of the tangent as though it were anyway. Draw a picture. You want a line tangent to y=x^2 that ALSO passes through (0,2). So pick any point on the curve. Say (x,x^2) and imagine a line passing through (x,x^2) and (0,2). What's the slope of that (just use m=delta(y)/delta(x)). Now that slope should be the same as the slope of the tangent to the curve at (x,x^2). As you've noted, that's 2*x. Equate those two slopes. You should get a quadratic equation to solve for x.
  5. Jul 2, 2008 #4
    I thought the point is on the curve but rather on the line, in which the line intersects that point for the first one.
  6. Jul 2, 2008 #5


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    You mean you thought the point is NOT on the curve? Yes, that is correct- the point (0,2) is not on the graph y= x2 but is on the tangent line. In your first post you were assuming that the point (2, 0) was on both the line and the curve- that it was the "point of tangency".
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