Question

Find an equation ion to the normal line to the graph of y=(3x^2+2x)^1/2 at (2,4) is

Answer 1

@zepdrix

Answer 2

Ugh?

Answer 3

Okay?

Answer 4

@mathmale

Answer 5

Do you remember how to find `tangent line`?

Answer 6

Take the derivative right?

Answer 7

Maybe o_O

Answer 8

at the point (2,4), f'(2) gives us the slope of the tangent line.

The slope of the normal line will be the `negative reciprocal of the tangent slope`.

Answer 9

\[\large\rm m=-\frac{1}{f'(2)}\]

Answer 10

So yes, start by finding a derivative.

Answer 11

Okay

Answer 12

Do you want it in square root form?

Answer 13

Well 1/2(6x+2)/square root 6x^2+2x

Answer 14

Omg :/ of course

Answer 15

She's here, her aunt usually takes 10 hours to shop

Answer 16

\[\large\rm f'(x)=\frac{1}{2}(3x^2+2x)^{-1/2}(6x+2)\]Ok good.
Evaluate at x=2.

Answer 17

oh lol :P ok peace out

Answer 18

We can just keep this and than if we're on at the same time one of us can tag eachother

Answer 19

ya

Answer 20

@zepdrix when you get on :D

Answer 21

\[\large\rm f'(x)=\frac{1}{2}(3x^2+2x)^{-1/2}(6x+2)\]Evaluate at x=2,\[\large\rm f'(2)=\frac{1}{2}(3(2)^2+2(2))^{-1/2}(6(2)+2)\]Simplify.
This is the tangent slope.
Flip it, negative it, this is the normal slope.

Answer 22

@zepdrix

Answer 23

7/4?

Answer 24

So -4/7? For normal line

Answer 25

Good good good, that is our `normal slope`.

Answer 26

\[\large\rm m=-\frac{4}{7}\]

Answer 27

When you've done these problems in the past,
do you usually set them up in `point-slope form` or `slope-intercept form`?

Answer 28

Point slope

Answer 29

Usually

Answer 30

Ok good, then we're done.

Answer 31

\(\large\rm y-y_o=m(x-x_o)\)
Just plug in the stuff.

Answer 32

\(\large\rm (x_o,y_o)=(2,4)\)\[\large\rm m=-\frac47\]

Answer 33

Y-4=-4/7(x-2)

Answer 34

yay team

Answer 35

:)

Answer 36

So?