If y = 9 and x = 3, then how to calculate the slope?

Indranil

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f(x) = x^2 and if the x = 3, then y = 9 and x = 3 in coordinate plane. As I know if we differentiate x^2 we get 2x, so f(x) = x^2 to f'(x) = 2x = 6 (after differentiating x^2). How to calculate the slope? Could you provide an diagram?
 
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f(x) = x^2 and if the x = 3, then y = 9 and x = 3 in coordinate plane. As I know if we differentiate x^2 we get 2x, so f(x) = x^2 to f'(x) = 2x = 6 (after differentiating x^2). How to calculate the slope? Could you provide an diagram?
Do you know the relationship between the derivative of a function (in 2D) at a point and the slope of the tangent line (to the curve made by the function) at that point?
 
f(x) = x^2 and if the x = 3, then y = 9 and x = 3 in coordinate plane. As I know if we differentiate x^2 we get 2x, so f(x) = x^2 to f'(x) = 2x = 6 (after differentiating x^2). How to calculate the slope? Could you provide an diagram?

draw the graph of f(x)=x2
then draw the slope of the curve (the line tangent to the curve) at x=2, x=3, etc.

the derivative is the eqn for the slope of the tangent line,
(or/ the slope of the curve at a given point x).
f'(x)=2x
at x=2, the slope=4,
at x=3, the slope=6, and so on.
 
… Could you provide [a] diagram?
Here's an animated graph.

You need to edit the function f(x) and its first derivative g(x), to get your parabola and its tangent lines.

Change f(x) to x^2

Change g(x) to 2x

You may zoom in and out. You may pause the animation, by dragging the blue slider dot to the value of x=a that you want.
 
Here's an animated graph.

You need to edit the function f(x) and its first derivative g(x), to get your parabola and its tangent lines.

Change f(x) to x^2

Change g(x) to 2x

You may zoom in and out. You may pause the animation, by dragging the blue slider dot to the value of x=a that you want.
I have tried but I am unable to draw the graph on the site you provided. What can I do now? Please help.
 
I have tried but I am unable to draw the graph …
I'm sorry you're having difficulties. I just created an account there, so that I could edit the functions f and g, for you.

This link works for me. It shows the tangent line at the point (3,9). If you would like to see the animation, the Play button is located on the far left side in box #5. :cool:
 
f(x) = x^2 and if the x = 3, then y = 9 and x = 3 in coordinate plane. As I know if we differentiate x^2 we get 2x, so f(x) = x^2 to f'(x) = 2x = 6 (after differentiating x^2). How to calculate the slope? Could you provide an diagram?
You should NOT write f'(x) = 2x = 6, as f'(x) is NOT 6. f'(x)=2x and f'(2)=6. You should try to learn what information f'(2)=6 gives you. This will help you figure out what the slope is.
 
Many, many years ago, when I was in high school, I was taught this method for drawing a tangent line to a graph:

Given the graph hold a small mirror (pocket or purse mirror) across the curve at the given point. Slowly rotate the mirror, at that point, until the image of the curve appears to go smoothly into its image in the mirror (no "corners" at the point). Use the mirror as a straight edge to draw a line. That line will be perpendicular to the given curve at that point (do you see why?).

Now do the same thing to draw the perpendicular to that perpendicular. That is, rotate the mirror slowly about that point until the perpendicular seems to go smoothly into its image. Again use the mirror as a straight edge to draw that line. That "perpendicular to the perpendicular to the curve" is tangent to the curve at that point.
 
Many, many years ago, when I was in high school, I was taught this method for drawing a tangent line to a graph:

Given the graph hold a small mirror (pocket or purse mirror) across the curve at the given point. Slowly rotate the mirror, at that point, until the image of the curve appears to go smoothly into its image in the mirror (no "corners" at the point). Use the mirror as a straight edge to draw a line. That line will be perpendicular to the given curve at that point (do you see why?).

Now do the same thing to draw the perpendicular to that perpendicular. That is, rotate the mirror slowly about that point until the perpendicular seems to go smoothly into its image. Again use the mirror as a straight edge to draw that line. That "perpendicular to the perpendicular to the curve" is tangent to the curve at that point.
Aahhhh ...... brought back memories.....
 
I have tried but I am unable to draw the graph on the site you provided. What can I do now? Please help.

draw the graph on a piece of paper.

you know one point on the tangent line; 3,9 =x2,y2
slope = 6=(y2-y1)/(x2-x1) will give you another,
i.e. find y1 for a value of x1, (let x1=0 is easiest)
 
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I'm sorry you're having difficulties. I just created an account there, so that I could edit the functions f and g, for you.

This link works for me. It shows the tangent line at the point (3,9). If you would like to see the animation, the Play button is located on the far left side in box #5. :cool:
I can't see the slop there. Could you point to the slop there, please?
 
I can't see the slop there.
Gosh. I took the time to create an account and provide the diagram that you had asked for, and now you call my graph slop? ;) heh, heh

The correct name is "slope".

You can't see the slope on a diagram. You need to calculate the slope. I had thought that you asked for a diagram to see what a tangent line looks like.

We already told you how to calculate the slope! Why don't you do it?

The first derivative of f(x) gives you the slope at each value of x.

You know the first derivative: f'(x) = 2x

For example, if I wanted to know the slope of the tangent line to the parabola at the point (4,16), I would use the derivative.

f'(4) = 8

The tangent line has slope 8, at the point (4, 16)

f'(-3/2) = -3

The tangent line has slope -3, at the point (-3/2, 9/4)

f'(0) = 0

The tangent line has slope 0, at the point (0, 0)

Maybe part of your difficulties arise from not knowing prerequisite material. You need experience with precalculus topics, before studying calculus.
 
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Gosh. I took the time to create an account and provide the diagram that you had asked for, and now you call my graph slop? ;) heh, heh

The correct name is "slope".

You can't see the slope on a diagram. You need to calculate the slope. I had thought that you asked for a diagram to see what a tangent line looks like.

We already told you how to calculate the slope! Why don't you do it?

The first derivative of f(x) gives you the slope at each value of x.

You know the first derivative: f'(x) = 2x

For example, if I wanted to know the slope of the tangent line to the parabola at the point (4,16), I would use the derivative.

f'(4) = 8

The tangent line has slope 8, at the point (4, 16)

f'(-3/2) = -3

The tangent line has slope -3, at the point (-3/2, 9/4)

f'(0) = 0

The tangent line has slope 0, at the point (0, 0)

Maybe part of your difficulties arise from not knowing prerequisite material. You need experience with precalculus topics, before studying calculus.
I appreciate you for doing the efforts for me that's why I shall be grateful to you forever. Thank you all for helping me all the way possible. Thanks a lot.
 
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