Few Problems

1. Jul 1, 2008

razored

I have multiple problems :

Find the tangent to the curve $$y=x^{2}$$ going through the point $$(2,0)$$. Note that $$(2,0)$$ 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 $$y=x^{2}$$ 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 $$y=x^{2} + bx +c$$have the line y=x as a tangent at the point with abscissa 2?

No clue.

2. Jul 1, 2008

HallsofIvy

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.)

3. Jul 1, 2008

Dick

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.

4. Jul 2, 2008

razored

I thought the point is on the curve but rather on the line, in which the line intersects that point for the first one.

5. Jul 2, 2008

HallsofIvy

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".