Find the equation of tangent and normal for $y=f(x)$, if, $ y=x^2-2x+3$ and tangent is perpendicular to line $x+y-1=0$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Cartesian Equation for the perpendicular bisector of a lineFind Equation of a Perpendicular Line Going Through a PointPerpendicular form of the straight line equation.Find equation of a line perpendicular to the tangent of curve at a given point.Find the equation of parabola tangent to a lineFind the equation of the straight line which is both a tangent and normal to the curve $x=3t^2,y=2t^3?$Equation of plane through a point and perpendicular to planesFind the tangentFind the condition that the line $lx+my+n=0$.Condition for a line to be tangent to a parabola and normal to another one simultaneously
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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 $x+y-1=0$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Cartesian Equation for the perpendicular bisector of a lineFind Equation of a Perpendicular Line Going Through a PointPerpendicular form of the straight line equation.Find equation of a line perpendicular to the tangent of curve at a given point.Find the equation of parabola tangent to a lineFind the equation of the straight line which is both a tangent and normal to the curve $x=3t^2,y=2t^3?$Equation of plane through a point and perpendicular to planesFind the tangentFind the condition that the line $lx+my+n=0$.Condition for a line to be tangent to a parabola and normal to another one simultaneously
$begingroup$
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
asked Mar 25 at 20:38
J. DoeJ. Doe
1
$endgroup$
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
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
asked Mar 25 at 20:38
J. DoeJ. Doe
1
$endgroup$
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
analytic-geometry
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
asked Mar 25 at 20:38
J. DoeJ. Doe
1
asked Mar 25 at 20:38
J. DoeJ. Doe
1
1
$begingroup$
Welcome to Math Stack Exchange. Looks correct to me
$endgroup$
– J. W. Tanner
Mar 25 at 20:42
add a comment |
$begingroup$
Welcome to Math Stack Exchange. Looks correct to me
$endgroup$
– J. W. Tanner
Mar 25 at 20:42
$begingroup$
Welcome to Math Stack Exchange. Looks correct to me
$endgroup$
– J. W. Tanner
Mar 25 at 20:42
$begingroup$
Welcome to Math Stack Exchange. Looks correct to me
$endgroup$
– J. W. Tanner
Mar 25 at 20:42
add a comment |
1 Answer
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$begingroup$
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 $f(x)$.
answered Mar 26 at 0:59
J. W. TannerJ. W. Tanner
4,7871420
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
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 $f(x)$.
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 $f(x)$.
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 $f(x)$.
answered Mar 26 at 0:59
J. W. TannerJ. W. Tanner
4,7871420
$endgroup$
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 $f(x)$.
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
answered Mar 26 at 0:59
J. W. TannerJ. W. Tanner
4,7871420
4,7871420
add a comment |
add a comment |
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Welcome to Math Stack Exchange. Looks correct to me
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– J. W. Tanner
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