# Math Help - Tangent to a curve

1. ## Tangent to a curve

Find the value of c for which the line $y=x+c$ is a tangent to the curve $y=x^2-5x+4$
thanks!

2. Originally Posted by BabyMilo
thanks!
do i dy/dx it? then dy/dx=0

then sub x into

to get y.

then y=x+c
to find c?

thanks!

3. If it is tangent to the curve, it will touch the curve only once.

Since they touch, they must be equal.

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

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

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

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

$36 - 16 + 4c = 0$

$20 + 4c = 0$

$4c = -20$

$c = -5$.

4. 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 $x + c = x^2 - 5x + 4$

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

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

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

$36 - 16 + 4c = 0$

$20 + 4c = 0$

$4c = -20$

$c = -5$.
stupid me.

5. 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.

6. Here is a method using the derivative.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

That seems simpler to me.

8. Ah, yes, I didn't spot that