Parabola Question

[fullzeka]

What is the equation of the normal perpendicular(right) to the line tangent to the parabola and passes through the point x = 1?

I hope the question is understandable because i had to use translation to translate the terms.

Thanks for the help.

 

[topsquark]

I'm assuming there's a typo. Your function is not a parabola! I'm guessing you meant:
[math]y = 2x^2 + 5x - 2[/math]
Hint 1: First find the derivative of y for x = 1. This is the slope, m, of the tangent line at x = 1.

Hint 2: If you know the slope of a line, how do you find the slope of the perpendicular to that line?

-Dan

 

[fullzeka]

Hey, there is no typo. It is how it is.This is a question from the mid-term exam of my university.I might have done a mistake calling it a parabola question though. By the way unfortunately my responses are waiting for an approval so i respond slowly.

 

[pka]

What is the equation of the normal perpendicular(right) to the line tangent to the parabola View attachment 19511 and passes through the point x = 1?
I hope the question is understandable because i had to use translation to translate the terms.

Like Dan I assume that it is \(2x^2+5x-2\). The it must to be a parabola. The point \((1,5)\)
From the translation it hard to know if this is from a calculus class?
At a point on the parabola where \(x=a\) then the slope of tangent is \(4a+5\) so the normal line there has slope \(\dfrac{-1}{4a+5}\).
Please post the answer you find.

 

[fullzeka]

Like Dan I assume that it is \(2x^2+5x-2\). The it must to be a parabola. The point \((1,5)\)
From the translation it hard to know if this is from a calculus class?
At a point on the parabola where \(x=a\) then the slope of tangent is \(4a+5\) so the normal line there has slope \(\dfrac{-1}{4a+5}\).
Please post the answer you find.

It is 2x^3 and no , i study economics and this question is from the class called Mathematics II ...

 

[topsquark]

It is 2x^3 and no , i study economics and this question is from the class called Mathematics II ...

Well, if it isn't a parabola then it isn't a parabola. The approach is the same. Find the slope of the tangent at x = 1, find the slope perpendicular to that, then use that slope to find the equation of the perpendicular line.

See how that goes and let us know.

-Dan

 

[pka]

It is 2x^3 and no , i study economics and this question is from the class called Mathematics II ...

So it is \(2x^3+5x-2\). The point \((1,5)\) on the graph.
From the translation it hard to know if this is from a calculus class?
At a point on the graph where \(x=a\) then the slope of tangent is \(6a^2+5\) so the normal line there has slope \(\dfrac{-1}{6a^2+5}\).
Please post the answer you find.

 

[JeffM]

Well, if it isn't a parabola then it isn't a parabola. The approach is the same. Find the values of f(1) and f'(1), find the slope perpendicular to that, then use that slope to find the equation of the perpendicular line.

See how that goes and let us know.

-Dan

To rephrase topsquark a bit, find f'(x), evaluate f(1) and f'(1), find the slope of the normal to the tangent of f(x) at f(1), and then find the linear equation g(x) with that slope and g(1) = f(1).

The name of that type of equation in English is a cubic.

 

[fullzeka]

How does it look ? @JeffM @pka @topsquark

 

[willyengland]

Looks good.
Perhaps you could transform it into the standard form y = m x + b.

 

[HallsofIvy]

What is the equation of the normal perpendicular(right) to the line tangent to the parabola View attachment 19511 and passes through the point x = 1?

I hope the question is understandable because i had to use translation to translate the terms.

Thanks for the help.

The problem I have with this si that "x= 1" is NOT a point! The others are assuming you mean that the normal was to go through the point on the graph where x= 1. That is, through (1, 2+ 5- 2)= (1, 5). Is that correct?