# find tangent lines

• Feb 6th 2010, 10:31 AM
bigwave
find tangent lines
given

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

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

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

the answers are: $\displaystyle y=2x-1$ and $\displaystyle y=4x-4$
• Feb 6th 2010, 11:08 AM
running-gag
Quote:

Originally Posted by bigwave
given

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

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

how are the equations of 2 tangent lines found

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

Hi

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

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

Originally Posted by bigwave
given

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

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

how are the equations of 2 tangent lines found

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

1. You have the parabolas:

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

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:

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

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

4. Both equations describe the same line. Thus

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

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

6. For confirmation: Draw the 2 parabolas and the tangents.
• Feb 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 $\displaystyle x_0$ is
$\displaystyle y = f'(x_0)(x-x_0) + f(x_0)$

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

where did you get abscissa is $\displaystyle x_0 = 1$
• Feb 7th 2010, 01:21 AM
HallsofIvy
Quote:

Originally Posted by bigwave
given

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

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

how are the equations of 2 tangent lines found

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

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, $\displaystyle (x_0, y_0)$ where that line is tangent to $\displaystyle y= x^2$ we must have $\displaystyle y_0= x_0^2= mx_0+ b$ and $\displaystyle y'= 2x_0= m$. At the point, $\displaystyle (x_1, y_1)$ where it is tangent to $\displaystyle x^2+ 6x- 5$, we must have $\displaystyle y_1= x_1^2+ 6x_1- 5= mx_1+ b$ and $\displaystyle y'= 2x_1+ 6= m$ so we have 4 equations to solve for m, b, $\displaystyle x_1$, and $\displaystyle x_2$
• Feb 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..