# Math Help - tangent line & normal line question

1. ## tangent line & normal line question

here's the problem
Find the equations of the tangent line & normal line to the graph of the equation at the indicated point..
____
y = √x+1; (3,2)

i got the slope and its
__1___
2√x+1

so i have to plug it in the tangent line & normal line formula which is
$
y - y1 = m(x-x1) - tangent line$

and
$
y - y1 = 1/m(x-x1) - normal line$

so its goin' to be:

y-2 = 1/2√x+1 (x-3)

___ ___
0 =1/2√x+1 + 1/6√x+1

pls help..thanx

2. Originally Posted by emjez15
here's the problem
Find the equations of the tangent line & normal line to the graph of the equation at the indicated point..
____
y = √x+1; (3,2)

i got the slope and its
__1___
2√x+1

so i have to plug it in the tangent line & normal line formula which is
$
y - y1 = m(x-x1) - tangent line$

and
$
y - y1 = {\bold{\color{red}\ -\ }}1/m(x-x1) - normal line$
<<<<you forgot the negative sign

so its goin' to be:

y-2 = 1/2√x+1 (x-3)

___ ___
0 =1/2√x+1 + 1/6√x+1

pls help..thanx
From $f(x)=\sqrt{x+1}$ you got $f'(x)=\dfrac1{2\sqrt{x+1}}$ . That's correct!

Therefore you know that the slope of the tangent in T(3, 2) is:

$f'(3)= \dfrac1{2\sqrt{3+1}} = \dfrac1{2\cdot 2} = \dfrac14$

Now plug in the coordinates of the point T and the value of the slope to get the equation of the tangent:

$y - 2 = \dfrac14 \cdot (x-3) ~~~~~~~ \leftarrow\ \text{tangent line}$

The slope of the normal is $m_n =-\ \dfrac1{\frac14} = -4$

Now plug in the coordinates of the point T and the value of the slope to get the equation of the normal:

$y - 2 = -4 \cdot (x-3) ~~~~~~~ \leftarrow\ \text{normal line}$

3. yah, thanx dude,
i got it ..

but i've got 1 more problem

how can i solve this

$y = 4/x2 ; (2,1)$

4. Originally Posted by emjez15
yah, thanx dude,
i got it ..

but i've got 1 more problem

how can i solve this

$y = 4/x2 ; (2,1)$
you mean $y = \frac{4}{x^2} = 4x^{-2}$ ?

$\frac{dy}{dx} = -8x^{-3} = -\frac{8}{x^3}$

now find the tangent and normal lines at the given point on the curve.

same drill as before.

5. ahhh,, i see, got it..thats why..^^
thanks sir..^^