Draw the graph of y = x2 and the tangent at x = 1.?
回答 (6)
2020-07-20 8:44 pm
y = x²
When x = 1: y = 1
dy/dx = 2x
Slope of the tangent at (x = 1)
= 2(1)
= 2
Equation of the tangent:
y - 1 = 2(x - 1)
y - 1 = 2x - 2
2x - y - 1 = 0
The graph is shown as follows:
When x = 1: y = 1
dy/dx = 2x
Slope of the tangent at (x = 1)
= 2(1)
= 2
Equation of the tangent:
y - 1 = 2(x - 1)
y - 1 = 2x - 2
2x - y - 1 = 0
The graph is shown as follows:
2020-07-20 8:45 pm
Question:
Are you saying that you are studying calculus (limits, derivatives...), yet, you can't draw the graph of y = x^2 ? Surely you can do that part on your own, no?
Are you saying that you are studying calculus (limits, derivatives...), yet, you can't draw the graph of y = x^2 ? Surely you can do that part on your own, no?
2020-07-20 11:35 pm
y=x^2=>y(1)=1
y '=2x=>y '(1)=2
according to these derived information
the equation of the tangent at x=1 is
y-1=2(x-1)=>y=2x-1
Then, ploy the graph of
y=x^2
&
y=2x-1
on the same graph paper.
y '=2x=>y '(1)=2
according to these derived information
the equation of the tangent at x=1 is
y-1=2(x-1)=>y=2x-1
Then, ploy the graph of
y=x^2
&
y=2x-1
on the same graph paper.
2020-07-20 8:42 pm
You probably meant y = x^2.
Let tangent be y = mx + c
When x = 1, y = 1, m = dy/dx = 2
1 = 2 + c
y = 2x – 1
https://www.wolframalpha.com/input/?i=y+%3D+2x+%E2%80%93+1%2C+y+%3D+x%5E2%2C+%28+x+from+-2+to+2%29
Let tangent be y = mx + c
When x = 1, y = 1, m = dy/dx = 2
1 = 2 + c
y = 2x – 1
https://www.wolframalpha.com/input/?i=y+%3D+2x+%E2%80%93+1%2C+y+%3D+x%5E2%2C+%28+x+from+-2+to+2%29
2020-07-20 8:35 pm
y = x2 = 2x is a straight line and the tangent is itself.
收錄日期: 2021-04-18 18:35:12
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200720122732AAzdOpP