# Tangent to a curve

• Dec 27th 2009, 12:50 AM
BabyMilo
Tangent to a curve
Quote:

Find the value of c for which the line $\displaystyle y=x+c$ is a tangent to the curve $\displaystyle y=x^2-5x+4$
thanks!
• Dec 27th 2009, 12:54 AM
BabyMilo
Quote:

Originally Posted by BabyMilo
thanks!

do i dy/dx it? then dy/dx=0

then sub x into
http://www.mathhelpforum.com/math-he...23de3734-1.gif

to get y.

then y=x+c
to find c?

thanks!
• Dec 27th 2009, 12:58 AM
Prove It
If it is tangent to the curve, it will touch the curve only once.

Since they touch, they must be equal.

So $\displaystyle x + c = x^2 - 5x + 4$

$\displaystyle x^2 - 6x + 4 - c = 0$

Since it only touches once, the discriminant must be 0.

So $\displaystyle \Delta = (-6)^2 - 4(1)(4 - c) = 0$

$\displaystyle 36 - 16 + 4c = 0$

$\displaystyle 20 + 4c = 0$

$\displaystyle 4c = -20$

$\displaystyle c = -5$.
• Dec 27th 2009, 01:23 AM
BabyMilo
Quote:

Originally Posted by Prove It
If it is tangent to the curve, it will touch the curve only once.

Since they touch, they must be equal.

So $\displaystyle x + c = x^2 - 5x + 4$

$\displaystyle x^2 - 6x + 4 - c = 0$

Since it only touches once, the discriminant must be 0.

So $\displaystyle \Delta = (-6)^2 - 4(1)(4 - c) = 0$

$\displaystyle 36 - 16 + 4c = 0$

$\displaystyle 20 + 4c = 0$

$\displaystyle 4c = -20$

$\displaystyle c = -5$.

• Dec 27th 2009, 02:20 AM
HallsofIvy
This was posted in the "Pre Calculus" section so Prove It used a method that did not require the derivative.

Since you do mention the derivative, no, the derivative is not 0 where the line y= x+ c is tangent to it. A line is tangent to a curve where its slope is the same as the derivative. y= x+ c has slope 1 so you are looking for a place where the derivative is 1, not 0.
• Dec 27th 2009, 03:43 AM
Bacterius
Here is a method using the derivative.

--------------------

The derivative of the function $\displaystyle f(x) = x^2 - 5x + 4$ is $\displaystyle f'(x) = 2x - 5$ (power rule on sum of functions).

The line $\displaystyle y = x + c$ has a slope equal to $\displaystyle 1$. Thus, you are looking for the point on the curve of $\displaystyle f(x)$ where the slope is equal to $\displaystyle 1$. So, you must solve $\displaystyle 2x - 5 = 1$ for $\displaystyle x$. Hmm, $\displaystyle x = 3$.

Say $\displaystyle a = 3$ (to make it less confusing). You know that the tangent to the curve of $\displaystyle f$ in a point of absciss $\displaystyle a$ is equal to :

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

Substitute : $\displaystyle y = (x - 3) \times 1 - 2$

That is : $\displaystyle y = x - 5$. Therefore $\displaystyle c = -5$.

:)
• Dec 27th 2009, 05:11 AM
HallsofIvy
Quote:

Originally Posted by Bacterius
Here is a method using the derivative.

--------------------

The derivative of the function $\displaystyle f(x) = x^2 - 5x + 4$ is $\displaystyle f'(x) = 2x - 5$ (power rule on sum of functions).

The line $\displaystyle y = x + c$ has a slope equal to $\displaystyle 1$. Thus, you are looking for the point on the curve of $\displaystyle f(x)$ where the slope is equal to $\displaystyle 1$. So, you must solve $\displaystyle 2x - 5 = 1$ for $\displaystyle x$. Hmm, $\displaystyle x = 3$.

Say $\displaystyle a = 3$ (to make it less confusing). You know that the tangent to the curve of $\displaystyle f$ in a point of absciss $\displaystyle a$ is equal to :

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

Substitute : $\displaystyle y = (x - 3) \times 1 - 2$

That is : $\displaystyle y = x - 5$. Therefore $\displaystyle c = -5$.

:)

Since you were already given that y= x+ c, y= 3+ c= $\displaystyle 3^2- 5(3)+ 4= 9- 15+ 4= -2$ so c= -2-3= -5.

That seems simpler to me.
• Dec 27th 2009, 12:51 PM
Bacterius
Ah, yes, I didn't spot that :)