Normal line

[Petrus]

Hello,
i know how to calculate normal line on point but i dont know how to do this problem.
"find an equation of the normal line o the parabola y=x^2-5x+4 tha is parallel to the line x-3y=5."

 

[Opalg]

Hello,
i know how to calculate normal line on point but i dont know how to do this problem.
"find an equation of the normal line o the parabola y=x^2-5x+4 tha is parallel to the line x-3y=5."

If the normal has slope $1/3$ then the tangent has slope $-3$. Can you find a point on the parabola where the tangent has slope $-3$?

 

[Petrus]

If the normal has slope $1/3$ then the tangent has slope $-3$. Can you find a point on the parabola where the tangent has slope $-3$?

Hello,
I dont understand where you get them all, could you possible tell me how you know that?

 

[Opalg]

could you possible tell me how you know that?

You are told that the normal line is parallel to the line $x-3y=5$, which you can write as $y = \frac13x-\frac53.$ That is a line with slope $\frac13.$ The tangent is perpendicular to the normal, and you should know that if a line has slope $m$, then the perpendicular line has slope $-1/m.$ So if the normal has slope $\frac13$ then the tangent has slope $-3$. Does that help?

 

[Petrus]

If I know that how shall I proceed? I can't make any progress.

 

[MarkFL]

Hello Petrus,

If you compute the derivative of the parabola $f(x)=x^2-5x+4$, equate that to -3 and solve for $x$, you will then be able to determine the point on the parabola at which the normal line will intersect it.

Then you will have determined a point $(x,f(x))$ and the slope $m=-3$, and you may then apply the point-slope formula to obtain the equation of the desired normal line.

Can you follow these steps, and show us your work?

 

[Petrus]

Hello Petrus,

If you compute the derivative of the parabola $f(x)=x^2-5x+4$, equate that to -3 and solve for $x$, you will then be able to determine the point on the parabola at which the normal line will intersect it.

Then you will have determined a point $(x,f(x))$ and the slope $m=-3$, and you may then apply the point-slope formula to obtain the equation of the desired normal line.

Can you follow these steps, and show us your work?

Hello Mark,
we start with calculate x
first i derivate the function $f(x)=x^2-5x+4$ and get $f'(x)=2x-5$ and then i set that =-3 and get $x=1$ and then i put $x=1$ on orginal function and get $y=0$ so we get the point slope formula $y-y1=m(x-x1)$ and get $y-0=-3(x-1)$ and then $y=3x-3$

 

[MarkFL]

I agree with your work until the very last step. You correctly wrote:

$y-0=-3(x-1)$

Now, try distributing the -3 on the right again...

 

[Petrus]

I agree with your work until the very last step. You correctly wrote:

$y-0=-3(x-1)$

Now, try distributing the -3 on the right again...

Hello Mark,
Sorry i forgot i had -3 i mean $y=-3x+3$

 

[MarkFL]

Yes!

Good work, Petrus! I appreciate that you read the suggestions, applied them and posted your work as a means of following through! This lets us know that you understood the suggestions.

 

[Petrus]

Yes!

Good work, Petrus! I appreciate that you read the suggestions, applied them and posted your work as a means of following through! This lets us know that you understood the suggestions.

Thank you Mark and the one who replied early! Now i understand how to do it thanks to the help! Have a nice day!

 

[Opalg]

$y=-3x+3$

Good work!

 

[MarkFL]

We are glad to help, and it is gratifying to hear that your understanding has increased!

This is our goal.