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Find the equation of tangent and normal for $y=f(x)$, if, $ y=x^2-2x+3$ and tangent is perpendicular to line $x+y-1=0$

0

1

$begingroup$

I am supposed to find the equation of tangent and normal for $y=f(x)$, if, $ y=x^2-2x+3$ and tangent is perpendicular to line $x+y-1=0$. My solution for tangent is $y-x-3/4=0$ and for normal is $y+x-15/4=0$. Is it correct? Thanks!

analytic-geometry

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asked Mar 25 at 20:38

J. DoeJ. Doe

1

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• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Math Stack Exchange. Looks correct to me
  $endgroup$
  – J. W. Tanner
  Mar 25 at 20:42
  

  
  
  
  

add a comment | 

0

1

$begingroup$

I am supposed to find the equation of tangent and normal for $y=f(x)$, if, $ y=x^2-2x+3$ and tangent is perpendicular to line $x+y-1=0$. My solution for tangent is $y-x-3/4=0$ and for normal is $y+x-15/4=0$. Is it correct? Thanks!

analytic-geometry

share|cite|improve this question

asked Mar 25 at 20:38

J. DoeJ. Doe

1

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Math Stack Exchange. Looks correct to me
  $endgroup$
  – J. W. Tanner
  Mar 25 at 20:42
  

  
  
  
  

add a comment | 

$begingroup$

I am supposed to find the equation of tangent and normal for $y=f(x)$, if, $ y=x^2-2x+3$ and tangent is perpendicular to line $x+y-1=0$. My solution for tangent is $y-x-3/4=0$ and for normal is $y+x-15/4=0$. Is it correct? Thanks!

analytic-geometry

share|cite|improve this question

asked Mar 25 at 20:38

J. DoeJ. Doe

1

$endgroup$

share|cite|improve this question

asked Mar 25 at 20:38

J. DoeJ. Doe

1

share|cite|improve this question

asked Mar 25 at 20:38

J. DoeJ. Doe

1

asked Mar 25 at 20:38

J. DoeJ. Doe

1

asked Mar 25 at 20:38

J. DoeJ. Doe

1

1 Answer
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0

$begingroup$

It's always good to check your work in math.

When you think you have an answer, go back and make sure it works to answer the question.

To check that $y-x-3/4=0$ is perpendicular to $x+y-1=0$,

put the equations in slope-intercept form $y=x+3/4$ and $y=-x+1$;

since the product of the slopes is $-1 $, these lines are perpendicular.

Likewise, the line $y=x+3/4$ is perpendicular to $y=-x+15/4.$

The latter pair of lines intersects when $x+3/4=-x+15/4$, i.e., $x=3/2.$

At that point $y=9/4.$ Furthermore, $f(3/2)=9/4$, so this line is indeed tangent to $f(x)$.

share|cite|improve this answer

answered Mar 26 at 0:59

J. W. TannerJ. W. Tanner

4,7871420

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0

$begingroup$

It's always good to check your work in math.

When you think you have an answer, go back and make sure it works to answer the question.

To check that $y-x-3/4=0$ is perpendicular to $x+y-1=0$,

put the equations in slope-intercept form $y=x+3/4$ and $y=-x+1$;

since the product of the slopes is $-1 $, these lines are perpendicular.

Likewise, the line $y=x+3/4$ is perpendicular to $y=-x+15/4.$

The latter pair of lines intersects when $x+3/4=-x+15/4$, i.e., $x=3/2.$

At that point $y=9/4.$ Furthermore, $f(3/2)=9/4$, so this line is indeed tangent to $f(x)$.

share|cite|improve this answer

answered Mar 26 at 0:59

J. W. TannerJ. W. Tanner

4,7871420

$endgroup$

add a comment | 

0

$begingroup$

It's always good to check your work in math.

When you think you have an answer, go back and make sure it works to answer the question.

To check that $y-x-3/4=0$ is perpendicular to $x+y-1=0$,

put the equations in slope-intercept form $y=x+3/4$ and $y=-x+1$;

since the product of the slopes is $-1 $, these lines are perpendicular.

Likewise, the line $y=x+3/4$ is perpendicular to $y=-x+15/4.$

The latter pair of lines intersects when $x+3/4=-x+15/4$, i.e., $x=3/2.$

At that point $y=9/4.$ Furthermore, $f(3/2)=9/4$, so this line is indeed tangent to $f(x)$.

share|cite|improve this answer

answered Mar 26 at 0:59

J. W. TannerJ. W. Tanner

4,7871420

$endgroup$

add a comment | 

$begingroup$

It's always good to check your work in math.

When you think you have an answer, go back and make sure it works to answer the question.

To check that $y-x-3/4=0$ is perpendicular to $x+y-1=0$,

put the equations in slope-intercept form $y=x+3/4$ and $y=-x+1$;

since the product of the slopes is $-1 $, these lines are perpendicular.

Likewise, the line $y=x+3/4$ is perpendicular to $y=-x+15/4.$

The latter pair of lines intersects when $x+3/4=-x+15/4$, i.e., $x=3/2.$

At that point $y=9/4.$ Furthermore, $f(3/2)=9/4$, so this line is indeed tangent to $f(x)$.

share|cite|improve this answer

answered Mar 26 at 0:59

J. W. TannerJ. W. Tanner

4,7871420

$endgroup$

share|cite|improve this answer

answered Mar 26 at 0:59

J. W. TannerJ. W. Tanner

4,7871420

answered Mar 26 at 0:59

J. W. TannerJ. W. Tanner

4,7871420

answered Mar 26 at 0:59

J. W. TannerJ. W. Tanner

4,7871420