Question

How can I find the slope after I find the derivative? For instance, f(x) = 2x^2 - 3x - 5, x = 0. I solved that question using the derivative formula and I got the answer 4x - 2. Now, how do I find the slope?

Answer 1

the derivative is the equation of the slope at any given point

Answer 2

y' = 4x - 2; when x=0, then y' = ??

Answer 3

It's confusing me, because it says the answer is m= -3

Answer 4

i see, you dint derive it correctly, check it out again

Answer 5

2x^2 - 3x - 5 ; what do we derive it to?

Answer 6

Compute the derivative of the given function and find the slope of the line that is tangent to its graph for the specified value of the independent variable.
f(x) = 2x^2-3x-5; x =0

Answer 7

and your derivative is?

Answer 8

Thats the question. I used the derivative formula and I got the answer 4x -2

Answer 9

-3x does not derive down to -2

Answer 10

wow, thank you! You just made me realize what I wasn't seeing. I see it for all the problems now. So once I get the answer. The slope is the derivative when I plug in the X

Answer 11

correct:

y' = 4x - 3 , at x=0; y' = slope = -3

Answer 12

Are you able to help me out in one more thing?

Answer 13

maybe, depends on if i know it or not :)

Answer 14

I need to find the tangent

Answer 15

the tangent, or the equation of the tangent line?

Answer 16

I have to compute the derivative of the given function then find the equation of the line that is tangent to its graph for the specified value x = c

Answer 17

do you recall the point slope form of a line? from algebra?

Answer 18

m = y - y1 = m(x - x1)

Answer 19

sorry I didnt men to put M at the front

Answer 20

yes, but not the first part:

y - yo = m(x-xo) lets work on this a little

y - yo = mx - mxo

y = mx -mxo + yo

we know m = y' = -3 when x=0 so lets fill those in

y = -3x +3(9) + yo

we get yo form the original equation:

y = 4x^2 -3x -5 ; at x=0

Answer 21

y0 = -5 in that case so lets finish it

y = -3x +3(0) - 5

the equation of the tangent line at x=0 is then

y = -3x -5

Answer 22

its either that, or they want the point slope form itself, depends on the material i spose

Answer 23

What's y0? a y then zero?

Answer 24

its a slip of the typing :) the o and 0 are so close on the keyboard ...

Answer 25

that and i also mistyped

y = -3x +3(9) + yo
^^^
shoulda been 3(0) , but i hit the wrong key again

Answer 26

Ok, I get it. THANK YOU :)

Answer 27

youre welcome, and good luck :)

Answer 28

HOLD ON, M is the slope and y?

Answer 29

m is traditionally slope, but m = y' so dont get hung up on the letters

Answer 30

the important thing is to recognize what it is your after, and then adapt your own method of getting there

Answer 31

Oh so mx is just the X they give me

Answer 32

It says compute the derivative of the given function and then find the equation of the line that is tangent to its graph for the specified value x=c.
f(x) = 2; c=13
another problem is: f(x) 7-2x; c=5

Answer 33

lets try to organise it this way:

\[y_o=2x_o^2-3x_o-5;\ at\ x_o=0,\ y_o=-5\]\[y_o' = 4x_o-3;\ at\ x_o=0,\ y_o'=-3\]
now form the tangent equation \[y-y_o=y_o'(0)(x-x_o)\]
\[y+5=-3(x-0)\]and arrange as needed

Answer 34

is that from the problem f(x) = 2x^2-3x-5; x=0 ?