Question

Does anyone know how to find slope???? Please help

Answer 1

equation
Find the slope of the tangent to the curve at the point specified.
x^2+xy^2=32 at (4,2)
Slope=?

Answer 2

I got 32 for answer and didn;t know if that was right

Answer 3

http://www.mathwarehouse.com/algebra/linear_equation/slope-of-a-line.php

Answer 4

is the answer .5

Answer 5

What course level is this for? You can find the derivative and sub in the values for the given point. The derivative is, by definition, the slope of the tangent to the curve.

Answer 6

pre calc

Answer 7

Ok. Scratch that then. It is not a simply derivative either...

What kind of curve/shape does the equation graph?

Answer 8

a curved graph

Answer 9

hmm ... I just looked again and I think it graphs a circle.

Answer 10

oh sorry I put it wrong into my calculator

Answer 11

no problem. Have a look at where that point is on the circle.

Answer 12

ok got it

Answer 13

Great!

Answer 14

oh wiat now my calculator says error and I can't get the final answer

Answer 15

can you by any chance tell me what you got my calculator is not working now it keeps saying error

Answer 16

are you still there mrleiss

Answer 17

Sorry, yes I'm still here. You won't use your calculator. You need to graph it and look at the point on the circle that they gave you. Sketch the tangent line; the slope should be relatively clear from the sketch or from where the point is on the circle.

Answer 18

ok

Answer 19

then what I am starting to get confused

Answer 20

Vertical lines have undefined slopes; Horizontal lines have slopes of zero. The graph y = x has a slope of 1 and y = -x has a slope of -1. The line that runs through the circle at the point should be parallel to one of these and so should have the same slope.

Answer 21

so then the slope is 1

Answer 22

I think it might be -1. Sketch both y = x and y = -x to make sure.

Answer 23

yu just redid it and I did it the wrong way sorry I got it now

Answer 24

so then the answer is -1

Answer 25

is that the answer

Answer 26

are you still there???????

Answer 27

Take a look at this graphical @ http://is.gd/ou9DOM

Answer 28

\[x^2+xy^2=32 \]\[2x+y^2+2xy'=0\] put \(x=4\) and \(y=2\) and solve for \(y'\)

Answer 29

I got 32 so then is the slope 32

Answer 30

\[2\times 4+2^2+2\times 4\times y'=0\] etc

Answer 31

seems unlikely

Answer 32

\[8+4+8y'=0\]
\[12+8y'=0\]
\[8y'=-12\]
\[y'=-\frac{3}{2}\]

Answer 33

ok so then yup I got that now I re put it in the calculator so then the slope is(-3/2)

Answer 34

oh no i didn't
that is right

Answer 35

wait I;m confused so that is the slope or no

Answer 36

yes, it is \(-\frac{3}{2}\)

Answer 37

ok got it thank you