[SOLVED] Find the value of constant k in tangent

**looi76** · April 26th 2008, 07:35 AM

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

Attempt:

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

$y = k - 2x$

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

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

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

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

**Moo** · April 26th 2008, 07:40 AM

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 :

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

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

**looi76** · April 26th 2008, 07:52 AM

Thanks Moo!

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

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

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

I still can't continue!

**Mathstud28** · April 26th 2008, 07:56 AM

Originally Posted by looi76

Thanks Moo!

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

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

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

I still can't continue!

$y'=2x-6$

start wit that

**looi76** · April 26th 2008, 08:03 AM

Originally Posted by Mathstud28

$y'=2x-6$

start wit that

Thanks Mathstud!

You meant that I need to differentiate first.

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

**Mathstud28** · April 26th 2008, 08:06 AM

Originally Posted by looi76

Thanks Mathstud!

You meant that I need to differentiate first.

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

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

does that help?

**looi76** · April 26th 2008, 08:07 AM

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 :

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

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

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

Is the deritave -20 or deritave means something else?

**Mathstud28** · April 26th 2008, 08:09 AM

Originally Posted by looi76

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

Is the deritave -20 or deritave means something else?

derivative means differentiate $f'(x)=\lim_{h\to{0}}\frac{f(x+h)-f(x)}{h}$ where $f'(x)$ is the derivative

**Mathstud28** · April 26th 2008, 08:10 AM

Originally Posted by looi76

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

Is the deritave -20 or deritave means something else?

You are thinking of the discrminant in the quadractic formula

**Moo** · April 26th 2008, 08:15 AM

Ok, so let's do it step by step :

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

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

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

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

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

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

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

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

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

Hence $k=\dots$

**ThePerfectHacker** · April 29th 2008, 07:52 PM

Originally Posted by looi76

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

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

Thus, $y=-2x+k$ is tangent to $y=x^2-6x+14$ if and only if $-2x+k = x^2 - 6x + 14$ has exactly one solution. That means, the quadradic $x^2 - 4x + (14-k) = 0$ has exactly one solution. That happens precisely when the discriminant vanishes, thus, $16 - 4(14-k) = 0$ .

**HiteshRulZZZ** · November 25th 2012, 10:23 PM

Originally Posted by looi76

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

Attempt:

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

$y = k - 2x$

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

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

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

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 <---- Here's your Answer!!!

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