Equation of perpendicular line to tangent lines
I'm completely stuck on a question about equation of perpendicular lines to tangent lines.
I figured out my tangent lines equations, I know graphically what I should get for my perpendicular lines (x=-1 and x=1) but how can I prove that mathematically?
Thanks a lot for your help
graphing-functions tangent-line
asked Dec 8 at 21:56
Mario
211
add a comment |
I'm completely stuck on a question about equation of perpendicular lines to tangent lines.
I figured out my tangent lines equations, I know graphically what I should get for my perpendicular lines (x=-1 and x=1) but how can I prove that mathematically?
Thanks a lot for your help
graphing-functions tangent-line
asked Dec 8 at 21:56
Mario
211
Try to post what you did from next time.
– mm-crj
Dec 8 at 22:29
add a comment |
I'm completely stuck on a question about equation of perpendicular lines to tangent lines.
I figured out my tangent lines equations, I know graphically what I should get for my perpendicular lines (x=-1 and x=1) but how can I prove that mathematically?
Thanks a lot for your help
graphing-functions tangent-line
asked Dec 8 at 21:56
Mario
211
I'm completely stuck on a question about equation of perpendicular lines to tangent lines.
I figured out my tangent lines equations, I know graphically what I should get for my perpendicular lines (x=-1 and x=1) but how can I prove that mathematically?
Thanks a lot for your help
graphing-functions tangent-line
graphing-functions tangent-line
asked Dec 8 at 21:56
Mario
211
asked Dec 8 at 21:56
Mario
211
asked Dec 8 at 21:56
Mario
211
asked Dec 8 at 21:56
Mario
211
asked Dec 8 at 21:56
Mario
211
211
Try to post what you did from next time.
– mm-crj
Dec 8 at 22:29
add a comment |
Try to post what you did from next time.
– mm-crj
Dec 8 at 22:29
Try to post what you did from next time.
– mm-crj
Dec 8 at 22:29
Try to post what you did from next time.
– mm-crj
Dec 8 at 22:29
add a comment |
2 Answers
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A horizontal tangent line must have a slope of $0$. So that will be a line in the form of $y = 0*x + c = c$.
So at which points on the curve $y=x^3 -3x + 1$ is the slope $0$?
Well the derivative is the formula for the slope. (Assuming the function is differentiable).
So we must solve for $y' =3x^2 - 3 = 0$. Let $x_0, x_1$ be the two solutions to that.
So the equations of your tangent lines will be: $y= (x_0^3 -3x + 1)$ and $y = (x_1^3 -3x_1 + 1)$
2) A line perpendicular would be completely vertical. So they will have the form $x = c$ for some constant. In this case those lines are at $x=x_0$ and $x=x_1$. So those are you equations.
answered Dec 8 at 22:18
fleablood
68.1k22684
add a comment |
The equation of the curve is given by $y=x^3-3x+1$ Therefore the slope of the tangent is given by:
$$dot{y}=3x^2-3$$
Where the tangent is horizontal is when the slope of the tangent is zero $implies dot{y}=0$ or, $x=pm 1$. The points where this happens are given by $(1,-1)$ and $(-1,3)$.
Now the when the tangent is horizontal the normal will be vertical $implies$ it will have a form $x=c$. At $(1,-1)$ it is $x=1$ and at $(-1,3)$ it is $x=-1$.
$$$$
answered Dec 8 at 22:28
mm-crj
382212
add a comment |
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2 Answers
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2 Answers
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A horizontal tangent line must have a slope of $0$. So that will be a line in the form of $y = 0*x + c = c$.
So at which points on the curve $y=x^3 -3x + 1$ is the slope $0$?
Well the derivative is the formula for the slope. (Assuming the function is differentiable).
So we must solve for $y' =3x^2 - 3 = 0$. Let $x_0, x_1$ be the two solutions to that.
So the equations of your tangent lines will be: $y= (x_0^3 -3x + 1)$ and $y = (x_1^3 -3x_1 + 1)$
2) A line perpendicular would be completely vertical. So they will have the form $x = c$ for some constant. In this case those lines are at $x=x_0$ and $x=x_1$. So those are you equations.
answered Dec 8 at 22:18
fleablood
68.1k22684
add a comment |
A horizontal tangent line must have a slope of $0$. So that will be a line in the form of $y = 0*x + c = c$.
So at which points on the curve $y=x^3 -3x + 1$ is the slope $0$?
Well the derivative is the formula for the slope. (Assuming the function is differentiable).
So we must solve for $y' =3x^2 - 3 = 0$. Let $x_0, x_1$ be the two solutions to that.
So the equations of your tangent lines will be: $y= (x_0^3 -3x + 1)$ and $y = (x_1^3 -3x_1 + 1)$
2) A line perpendicular would be completely vertical. So they will have the form $x = c$ for some constant. In this case those lines are at $x=x_0$ and $x=x_1$. So those are you equations.
answered Dec 8 at 22:18
fleablood
68.1k22684
add a comment |
A horizontal tangent line must have a slope of $0$. So that will be a line in the form of $y = 0*x + c = c$.
So at which points on the curve $y=x^3 -3x + 1$ is the slope $0$?
Well the derivative is the formula for the slope. (Assuming the function is differentiable).
So we must solve for $y' =3x^2 - 3 = 0$. Let $x_0, x_1$ be the two solutions to that.
So the equations of your tangent lines will be: $y= (x_0^3 -3x + 1)$ and $y = (x_1^3 -3x_1 + 1)$
2) A line perpendicular would be completely vertical. So they will have the form $x = c$ for some constant. In this case those lines are at $x=x_0$ and $x=x_1$. So those are you equations.
answered Dec 8 at 22:18
fleablood
68.1k22684
A horizontal tangent line must have a slope of $0$. So that will be a line in the form of $y = 0*x + c = c$.
So at which points on the curve $y=x^3 -3x + 1$ is the slope $0$?
Well the derivative is the formula for the slope. (Assuming the function is differentiable).
So we must solve for $y' =3x^2 - 3 = 0$. Let $x_0, x_1$ be the two solutions to that.
So the equations of your tangent lines will be: $y= (x_0^3 -3x + 1)$ and $y = (x_1^3 -3x_1 + 1)$
2) A line perpendicular would be completely vertical. So they will have the form $x = c$ for some constant. In this case those lines are at $x=x_0$ and $x=x_1$. So those are you equations.
answered Dec 8 at 22:18
fleablood
68.1k22684
answered Dec 8 at 22:18
fleablood
68.1k22684
answered Dec 8 at 22:18
fleablood
68.1k22684
answered Dec 8 at 22:18
fleablood
68.1k22684
68.1k22684
add a comment |
add a comment |
The equation of the curve is given by $y=x^3-3x+1$ Therefore the slope of the tangent is given by:
$$dot{y}=3x^2-3$$
Where the tangent is horizontal is when the slope of the tangent is zero $implies dot{y}=0$ or, $x=pm 1$. The points where this happens are given by $(1,-1)$ and $(-1,3)$.
Now the when the tangent is horizontal the normal will be vertical $implies$ it will have a form $x=c$. At $(1,-1)$ it is $x=1$ and at $(-1,3)$ it is $x=-1$.
$$$$
answered Dec 8 at 22:28
mm-crj
382212
add a comment |
The equation of the curve is given by $y=x^3-3x+1$ Therefore the slope of the tangent is given by:
$$dot{y}=3x^2-3$$
Where the tangent is horizontal is when the slope of the tangent is zero $implies dot{y}=0$ or, $x=pm 1$. The points where this happens are given by $(1,-1)$ and $(-1,3)$.
Now the when the tangent is horizontal the normal will be vertical $implies$ it will have a form $x=c$. At $(1,-1)$ it is $x=1$ and at $(-1,3)$ it is $x=-1$.
$$$$
answered Dec 8 at 22:28
mm-crj
382212
add a comment |
The equation of the curve is given by $y=x^3-3x+1$ Therefore the slope of the tangent is given by:
$$dot{y}=3x^2-3$$
Where the tangent is horizontal is when the slope of the tangent is zero $implies dot{y}=0$ or, $x=pm 1$. The points where this happens are given by $(1,-1)$ and $(-1,3)$.
Now the when the tangent is horizontal the normal will be vertical $implies$ it will have a form $x=c$. At $(1,-1)$ it is $x=1$ and at $(-1,3)$ it is $x=-1$.
$$$$
answered Dec 8 at 22:28
mm-crj
382212
The equation of the curve is given by $y=x^3-3x+1$ Therefore the slope of the tangent is given by:
$$dot{y}=3x^2-3$$
Where the tangent is horizontal is when the slope of the tangent is zero $implies dot{y}=0$ or, $x=pm 1$. The points where this happens are given by $(1,-1)$ and $(-1,3)$.
Now the when the tangent is horizontal the normal will be vertical $implies$ it will have a form $x=c$. At $(1,-1)$ it is $x=1$ and at $(-1,3)$ it is $x=-1$.
$$$$
answered Dec 8 at 22:28
mm-crj
382212
answered Dec 8 at 22:28
mm-crj
382212
answered Dec 8 at 22:28
mm-crj
382212
answered Dec 8 at 22:28
mm-crj
382212
382212
add a comment |
add a comment |
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Try to post what you did from next time.
– mm-crj
Dec 8 at 22:29