Find another point on the curve with the same tangent.

**RogueDemon** · February 20th 2011, 06:54 AM

Question:

Find the equation of the tangent to $y = 8x^4 + 40x^3 + 26x^2 - 63x + 24$ at the point (3,-15). Find another point on the curve which has exactly the same equation for the tangent at the point.

Answer:

$f'(x) = 32x^3 + 120x^2 + 52x - 63$
$f'(-3) = m$
$m = -3$

$y - y_1 = m(x - x_1)$
$y = -3x + 6$

Now that I have the equation of the tangent, how do I go about finding the point? Do I have to set the equation of the tangent equal to the original equation and solve for $x$ , or is there a simpler / more correct way of doing such?

**TheEmptySet** · February 20th 2011, 07:17 AM

Originally Posted by RogueDemon

Question:

Find the equation of the tangent to $y = 8x^4 + 40x^3 + 26x^2 - 63x + 24$ at the point (3,-15). Find another point on the curve which has exactly the same equation for the tangent at the point.

Answer:

$f'(x) = 32x^3 + 120x^2 + 52x - 63$
$f'(-3) = m$
$m = -3$

$y - y_1 = m(x - x_1)$
$y = -3x + 6$

Now that I have the equation of the tangent, how do I go about finding the point? Do I have to set the equation of the tangent equal to the original equation and solve for $x$ , or is there a simpler / more correct way of doing such?

To have the same tangent line, the graph must have the same slope at that points. This gives $f'(x)=-3$

$f'(x)=32x^3+120x^2+52x-63=-3 \iff 32x^3+120x^2+52x-60=0$

Since you already know that $x=-3$ is a zero you can divide the cubic by $(x+3)$ to get

$(x+3)(32x^2+24x-20)$

The quadratic factor will give the possibilities for the x coordinate of the possible points. Can you finish from here?

**RogueDemon** · February 20th 2011, 07:25 AM

Yes, thanks very much!

**HallsofIvy** · February 20th 2011, 07:31 AM

Originally Posted by RogueDemon

Question:

Find the equation of the tangent to $y = 8x^4 + 40x^3 + 26x^2 - 63x + 24$ at the point (3,-15). Find another point on the curve which has exactly the same equation for the tangent at the point.

(3, -15) is not on the curve. I assume you mean (-3, 15).

Answer:

$f'(x) = 32x^3 + 120x^2 + 52x - 63$
$f'(-3) = m$

Okay, here you are using x= -3 so the (3, -15) was a typo.

$m = -3$

$y - y_1 = m(x - x_1)$

Better to write this out completely: y- (-15)= -3(x- 3) gives y+ 15= -3x+ 9 or y= -3x+ 6.

$y = -3x + 6$

Now that I have the equation of the tangent, how do I go about finding the point? Do I have to set the equation of the tangent equal to the original equation and solve for $x$ , or is there a simpler / more correct way of doing such?

You have to show that there is another x where that same tangent line is tangent to the curve. That means both that the values are the same and that they have the same slope there. Specifically, you need to find an x such that
$8x^4+ 40x^3+ 26x^2- 63x+ 24= -3x+ 6$
and $32x^3+ 120x^2- 52x- 63= -3$ which is the same as
$32x^3+ 120x^2- 52x- 60= 0$ .

Since you already know that x= -3 is a solution to the second equation, you can divide by x+ 3 to get a quadratic equation for two more values of x where the slope is -3. Check those in the first equation to see if one of them gives a point where the line and curve intersect.

**Soroban** · February 20th 2011, 08:13 AM

Hello, RogueDemon!

There is something wrong with the problem . . . please check.

$\text{Find the equation of the tangent to:}$
. . $y \:=\: 8x^4 + 40x^3 + 26x^2 - 63x + 24\,\text{ at the point }(3,\text{-}15).$

$\text{Find another point on the curve which has exactly the same equation}$
. . $\text{for the tangent at the point.}$

The point $(3,\text{-}15)$ is not on the curve.

. . The point $(3,1797)$ is on the curve.

. . The point $(\text{-}3,15)$ is on the curve.

What's going on?

Thread: Find another point on the curve with the same tangent.

LinkBack

Thread Tools

Display

Find another point on the curve with the same tangent.

Similar Math Help Forum Discussions

find each point at which the tangent line to the curve...

Find the Lines that are tangent and normal to curve at given point.

Find an equation of the tangent line to the curve at a given point.

Find the equation of the tangent line to the following curve at the given point P. Il

Find an equation of the tangent line to the curve at the point (7, 2).

Search Tags