# [SOLVED] Find the value of constant k in tangent

• Apr 26th 2008, 07:35 AM
looi76
[SOLVED] Find the value of constant k in tangent
Question:
Find the value of the constant $\displaystyle k$ for which the line $\displaystyle y + 2x = k$ is a tangent to the curve $\displaystyle y = x^2 - 6x + 14.$

Attempt:

$\displaystyle y = x^2 - 6x + 14$
$\displaystyle y + 2x = k$

$\displaystyle y = k - 2x$

$\displaystyle = x^2 - 6x + 2x + 14 - k$
$\displaystyle = x^2 - 4x + 14 - k$
$\displaystyle a = 1$, $\displaystyle b = -4$, $\displaystyle c = 14$

$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$\displaystyle x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4\times 1\times 14}}{2\times 1}$

I can't continue because Im getting a negative value inside the square root, where did I go wrong?
• Apr 26th 2008, 07:40 AM
Moo
Hello,

They're not saying that y+2x=k is the equation of the curve, but that it is tangent to the curve.

This means that at a given point, the tangent to the curve will have y+2x=k as an equation.

The tangent to the curve at a point of absciss a is :

$\displaystyle y=f'(a)(x-a)+f(a)$

Hence, calculate the derivative of $\displaystyle y = x^2 - 6x + 14$ and then... think :D
• Apr 26th 2008, 07:52 AM
looi76
Thanks Moo!

$\displaystyle y = x^2 - 6x + 14$
$\displaystyle a = 1$, $\displaystyle b = -6$, $\displaystyle c = 14$

$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$\displaystyle x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4\times1\times14}}{2\times1}$

I still can't continue!
• Apr 26th 2008, 07:56 AM
Mathstud28
Quote:

Originally Posted by looi76
Thanks Moo!

$\displaystyle y = x^2 - 6x + 14$
$\displaystyle a = 1$, $\displaystyle b = -6$, $\displaystyle c = 14$

$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$\displaystyle x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4\times1\times14}}{2\times1}$

I still can't continue!

$\displaystyle y'=2x-6$

start wit that
• Apr 26th 2008, 08:03 AM
looi76
Quote:

Originally Posted by Mathstud28
$\displaystyle y'=2x-6$

start wit that

Thanks Mathstud!

You meant that I need to differentiate first.

After I differentiate and get $\displaystyle y = 2x - 6$. What should I do next?
• Apr 26th 2008, 08:06 AM
Mathstud28
Quote:

Originally Posted by looi76
Thanks Mathstud!

You meant that I need to differentiate first.

After I differentiate and get $\displaystyle y = 2x - 6$. What should I do next?

If $\displaystyle y=-2x+k$ is to be a tangent line to $\displaystyle y=x^2-6x+14$..it must satify $\displaystyle =y-f(x_0)=f'(x_0)(x-x_0)$ Like Moo correctly stated...therefore we get $\displaystyle y-f(x_0)=(2(x_0)-6)(x-x_0)$

does that help?
• Apr 26th 2008, 08:07 AM
looi76
Quote:

Originally Posted by Moo
Hello,

They're not saying that y+2x=k is the equation of the curve, but that it is tangent to the curve.

This means that at a given point, the tangent to the curve will have y+2x=k as an equation.

The tangent to the curve at a point of absciss a is :

$\displaystyle y=f'(a)(x-a)+f(a)$

Hence, calculate the derivative of $\displaystyle y = x^2 - 6x + 14$ and then... think :D

The deritave of $\displaystyle y = x^2 - 6x + 14$.
$\displaystyle = b^2 - 4ac$
$\displaystyle = (-6)^2 - 4\times1\times14$
$\displaystyle = -20$

Is the deritave -20 or deritave means something else?
• Apr 26th 2008, 08:09 AM
Mathstud28
Quote:

Originally Posted by looi76
The deritave of $\displaystyle y = x^2 - 6x + 14$.
$\displaystyle = b^2 - 4ac$
$\displaystyle = (-6)^2 - 4\times1\times14$
$\displaystyle = -20$

Is the deritave -20 or deritave means something else?

derivative means differentiate $\displaystyle f'(x)=\lim_{h\to{0}}\frac{f(x+h)-f(x)}{h}$ where $\displaystyle f'(x)$ is the derivative
• Apr 26th 2008, 08:10 AM
Mathstud28
Quote:

Originally Posted by looi76
The deritave of $\displaystyle y = x^2 - 6x + 14$.
$\displaystyle = b^2 - 4ac$
$\displaystyle = (-6)^2 - 4\times1\times14$
$\displaystyle = -20$

Is the deritave -20 or deritave means something else?

You are thinking of the discrminant in the quadractic formula
• Apr 26th 2008, 08:15 AM
Moo
Ok, so let's do it step by step :

Let $\displaystyle f(x)=x^2 - 6x + 14$

$\displaystyle f'(x)=2x-6 \Longrightarrow f'(a)=2a-6$

Hence the equation of the tangent line at a point is :

$\displaystyle y=(2a-6)(x-a)+f(a)$

$\displaystyle y=(2a-6)(x-a)+a^2-6a+14$

$\displaystyle y=2ax-2a^2-6x+6a+a^2-6a+14$

$\displaystyle y=2ax-a^2-6x+14$

$\displaystyle y+(6-2a)x=14-a^2$

So the only point where there will be a tangent line whose equation is in the form $\displaystyle y+2x=k$ will be for $\displaystyle 6-2a=2 \Longleftrightarrow a=\dots$

Hence $\displaystyle k=\dots$

:)
• Apr 29th 2008, 07:52 PM
ThePerfectHacker
Quote:

Originally Posted by looi76
Question:
Find the value of the constant $\displaystyle k$ for which the line $\displaystyle y + 2x = k$ is a tangent to the curve $\displaystyle y = x^2 - 6x + 14.$

In the case of a parabola a line is tangent if and only if it has one intersection point with the parabola.

Thus, $\displaystyle y=-2x+k$ is tangent to $\displaystyle y=x^2-6x+14$ if and only if $\displaystyle -2x+k = x^2 - 6x + 14$ has exactly one solution. That means, the quadradic $\displaystyle x^2 - 4x + (14-k) = 0$ has exactly one solution. That happens precisely when the discriminant vanishes, thus, $\displaystyle 16 - 4(14-k) = 0$.
• Nov 25th 2012, 10:23 PM
HiteshRulZZZ
Re: [SOLVED] Find the value of constant k in tangent
Quote:

Originally Posted by looi76
Question:
Find the value of the constant $\displaystyle k$ for which the line $\displaystyle y + 2x = k$ is a tangent to the curve $\displaystyle y = x^2 - 6x + 14.$

Attempt:

$\displaystyle y = x^2 - 6x + 14$
$\displaystyle y + 2x = k$

$\displaystyle y = k - 2x$

$\displaystyle = x^2 - 6x + 2x + 14 - k$
$\displaystyle = x^2 - 4x + 14 - k$
$\displaystyle a = 1$, $\displaystyle b = -4$, $\displaystyle c = 14$

$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$\displaystyle x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4\times 1\times 14}}{2\times 1}$

I can't continue because Im getting a negative value inside the square root, where did I go wrong?

Here is how you do it.. Easy and Simple

Line: y = k-2x
Curve: y = x²-6x+14

Workings

k-2x = x²-6x+14

x²-6x+2x+14-k = 0

x²-4x+14-k = 0

a=1, b=-4, c=(14-k)

[I]Use this Equation[/I] ---- b²-4ac

b²-4ac
Replace..

=(-4)²-[4(14-k)]

=16-(56-4k)

=16-56+4k

4k = -16+56

4k = 40

k = 40/4

k = 10