find tangent lines

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February 6th 2010, 10:31 AM
bigwave

find tangent lines

given

$y=x^2$ and $y=-x^2+6x-5$

I know the derivatives are $2x$ and $-2x+6$

how are the equations of 2 tangent lines found to both graphs

the answers are: $y=2x-1$ and $y=4x-4$
February 6th 2010, 11:08 AM
running-gag

Quote:

Originally Posted by bigwave

given

$y=x^2$ and $y=-x^2+6x-5$

I know the derivatives are $2x$ and $-2x+6$

how are the equations of 2 tangent lines found

the answers are: $y=2x-1$ and $y=4x-4$

Hi

The equation of the tangent line at a point whose abscissa is $x_0$ is
$y = f'(x_0)(x-x_0) + f(x_0)$

$f(x)=x^2$
$f'(x)=2x$
The equation of the tangent line at the point whose abscissa is $x_0 = 1$ is
$y = f'(1)(x-1) + f(1)$
$y = 2(x-1) + 1$
$y = 2x-1$
February 6th 2010, 11:12 AM
earboth

Quote:

Originally Posted by bigwave

given

$y=x^2$ and $y=-x^2+6x-5$

I know the derivatives are $2x$ and $-2x+6$

how are the equations of 2 tangent lines found

the answers are: $y=2x-1$ and $y=4x-4$

1. You have the parabolas:

$p: y = x^2$
and
$q: y = -x^2+6x-5$

2. Let P(p, p²) denote the tangent point on the parabola p and Q(q, -q²+6q-5) the tangent point on the parabola q.

3. The tangent to p at P has the equation:

$t_p: y = 2px - p^2$
and the tangent to q at Q has the equation

$t_q: y = (-2q+6)x +q^2-5$

4. Both equations describe the same line. Thus

$\left|\begin{array}{rcl}2p&=&-2q+6 \\p^2+q^2&=&5 \end{array}\right.$

5. Solve this system of simultaneous equations for p and q. Resubstitute the result into the equations of $t_p$ and $t_q$ .

6. For confirmation: Draw the 2 parabolas and the tangents.
February 6th 2010, 02:24 PM
bigwave

Quote:

Originally Posted by running-gag

Hi

The equation of the tangent line at a point whose abscissa is $x_0$ is
$y = f'(x_0)(x-x_0) + f(x_0)$

$f(x)=x^2$
$f'(x)=2x$
The equation of the tangent line at the point whose abscissa is $x_0 = 1$ is
$y = f'(1)(x-1) + f(1)$
$y = 2(x-1) + 1$
$y = 2x-1$

where did you get abscissa is $x_0 = 1$
February 7th 2010, 01:21 AM
HallsofIvy

Quote:

Originally Posted by bigwave

given

$y=x^2$ and $y=-x^2+6x-5$

I know the derivatives are $2x$ and $-2x+6$

how are the equations of 2 tangent lines found

the answers are: $y=2x-1$ and $y=4x-4$

Here's a tip- in the future tell us what the problem really says! Here, you have never said exactly which tangent lines you want to find. Every graph has an infinite number of tangent lines.

I suspect that the problem is asking for lines that at tangent to both of these graphs. Suppose y= mx+ b is the equation of such a tangent line. At the point, $(x_0, y_0)$ where that line is tangent to $y= x^2$ we must have $y_0= x_0^2= mx_0+ b$ and $y'= 2x_0= m$ . At the point, $(x_1, y_1)$ where it is tangent to $x^2+ 6x- 5$ , we must have $y_1= x_1^2+ 6x_1- 5= mx_1+ b$ and $y'= 2x_1+ 6= m$ so we have 4 equations to solve for m, b, $x_1$ , and $x_2$
February 7th 2010, 08:12 PM
bigwave

yes the 2 tangent lines would be tangent to both graphs.

it was baffeling to determine the slope of the 2 lines just based on the derivatives and not knowing the points of contact..

appreciate the replys