LinusCalcTips

[Arty]

This question makes 0 sense to me.

can someone explain how its done.?

"explain why there is no value of k for which the graph of y=x^2+kx is tangent to the line y=5"

 

 

 

Fixed error in question 

Edited September 9, 2015 by Arty

[manikyath]

EDIT: NVM, got my terminology wrong, i'm a derp.

[wardude1]

so this means ks can't be 5?

[Arty]

so this means ks can't be 5?

typo 

it should be

"explain why there is no value of k for which the graph of y=x^2+kx is tangent to the line y=5"

[manikyath]

typo 

it should be

"explain why there is no value of k for which the graph of y=x^2+kx is tangent to the line y=5"

because in no instance of this the horizontal bit of the function willb get above zero.

 

in this case k will make the horizontal bit move down left if positive, and down right if negative.

 

EDIT: get wzgrapher, its awesome. its kinda dodgy to find tho...

[wardude1]

If i understood this correctly it is pretty darm simple

 

there is only one place where a function of x^2 can be tangent to a horizontal line (the bottom or top from the function) (i dont know exactly what tangent means (language barier))

this accures when you fill in x=0. if you do so the function will become 0 because 0^2 + 0k= 0 whatever you pick for k

 

hope you will understeand

[colonel_mortis]

You need do differentiate y=x²+kx, which gives you dy/dx=2x+k. If you set that to 0 (because the gradient of y=5 is 0), you get that k=-2x. Substitute that back into the original equation to get y=x²-2x²=-x². -x² can never be positive because x² must always be positive (assuming you're using real numbers), so it can't be tangential to y=5.

 

I feel like there was an easier way to do it, but that's the way I proved it just working through it in this reply box.

[wardude1]

You need do differentiate y=x²+kx, which gives you dy/dx=2x+k. If you set that to 0 (because the gradient of y=5 is 0), you get that k=-2x. Substitute that back into the original equation to get y=x²-2x²=-x². -x² can never be positive because x² must always be positive (assuming you're using real numbers), so it can't be tangential to y=5.

 

I feel like there was an easier way to do it, but that's the way I proved it just working through it in this reply box.

this is how you should make this one  :wacko:

[manikyath]

You need do differentiate y=x²+kx, which gives you dy/dx=2x+k. If you set that to 0 (because the gradient of y=5 is 0), you get that k=-2x. Substitute that back into the original equation to get y=x²-2x²=-x². -x² can never be positive because x² must always be positive (assuming you're using real numbers), so it can't be tangential to y=5.

 

I feel like there was an easier way to do it, but that's the way I proved it just working through it in this reply box.

this stuff is why i failed math class.

 

i lied, i miraculously passed by guessing...

[Knaj]

Well if you google x^2 it makes a weird "U" so that solves one part.

 

And if you google KX It's a lime motorbike

 

So, in conclusion

 

Y = U riding a motorbike

 

So you're riding a motorbike while figuring out Y

 

[wardude1]

this stuff is why i failed math class.

 

i lied, i miraculously passed by guessing...

I wanted to make a belgium joke but then i decided it would been lame

[Rolling Potatoe]

simple explanation:

 

the graph can never be a tangent to the y=5 line, no matter what value k has.

[manikyath]

I wanted to make a belgium joke but then i decided it would been lame

dont worry, we make an equal amount of fun of you guys.

[Arty]

If i understood this correctly it is pretty darm simple

 

there is only one place where a function of x^2 can be tangent to a horizontal line (the bottom or top from the function) (i dont know exactly what tangent means (language barier))

this accures when you fill in x=0. if you do so the function will become 0 because 0^2 + 0k= 0 whatever you pick for k

 

hope you will understeand

 

 

You need do differentiate y=x²+kx, which gives you dy/dx=2x+k. If you set that to 0 (because the gradient of y=5 is 0), you get that k=-2x. Substitute that back into the original equation to get y=x²-2x²=-x². -x² can never be positive because x² must always be positive (assuming you're using real numbers), so it can't be tangential to y=5.

 

I feel like there was an easier way to do it, but that's the way I proved it just working through it in this reply box.

Thanks to all

 

because in no instance of this the horizontal bit of the function willb get above zero.

 

in this case k will make the horizontal bit move down left if positive, and down right if negative.

 

EDIT: get wzgrapher, its awesome. its kinda dodgy to find tho...

 

 

so this means ks can't be 5?

THIS righ here 

i was trying to make those 2 equations tangent.........woops

 

 

 

 

 

 

 

 

 

 

this is how you should make this one  :wacko:

 

 

this stuff is why i failed math class.

 

i lied, i miraculously passed by guessing...

[wardude1]

dont worry, we make an equal amount of fun of you guys.

yeah i know and my dad is a vlamische person

[RazerZ]

Takes about 30 seconds to solve it.

 

Set the equations equal to each other and solve it. You now have a single quadratic equation. Now find the roots of the equation, if it has any. Using your knowledge of discriminants you will see that the equation must have two real roots. That means it will intersect twice instead of a single time, so it can't be tangent. 

 

y=x²+kx=5

 

x²+kx-5=0

 

b²-4ac .....  a=1, b=kx, c=-5

 

(kx)²-4(1)(5)

 

[Discriminants of a quadratic]

b²-4ac<0 no real roots

b²-4ac=0 one real root

b²-4ac>0 two real roots

 

(kx)²+20

 

No matter what value k is, your end result will always be positive since the term is being squared and you are adding a positive constant (20).

 

If you don't fully understand the problem, never be afraid to ask for help.

[Arty]

 

 

If you don't fully understand the problem, never be afraid to ask for help.

 

, alr

 

my hardest part was just understanding what it was asking

[Nineshadow]

@Arty

@colonel_mortis Here's a simpler way:

 

f_k(x)=x^2+kx is a quadratic function.

 

We need to prove that whatever k , the image of the function is tangent to y=5.

A quadratic function has the general form of : ax^2 + bx + c.

In our case :

a=1

b=k

c=0

a>0 , which means the parabola opens upward.

Let's calculate the discriminant. I'll replace delta with D.

D = b^2 - 4ac = k^2

 

k^2 is always bigger or equal to 0, this means we will always have 2 real solutions to our equation (x^2 + kx=0). The solutions will obviously intersect the line of y=0.

Somewhere between those solutions, we can find the vertex of the function, the point where the function "turns around".

 

For our parabola to be tangent to the line y=5, only one point needs to intersect that line. This means that the vertex of the function must be found somewhere on the y=5 line.

We have a formula for the positions of the vertex:

So, our problem becomes proving that -D/4a =5

D=k^2

a=1

-k^2/4=5

-1/4 * k^2 - 5 = 0

This is another quadratic function with a = -1/4 , b = 0, c = -5

We calculate the discriminant of our new equation :

D = b^2 - 4ac = - 4 * (-5) * (-1/4) =... < 0

We can see that the discriminant is smaller than zero (there's three negative numbers multiplied, resulting in a negative number). This means that the equation has no real solutions, thus proving our initial theory wrong.

 

The function x^2 - kx can't be tangent to the line y=5.

 

You don't need some things I explained, but it's good to set your basis when you work with a function , you never know when you will need some stuff

[Arty]

@Arty

f_k(x)=x^2+kx is a quadratic function.

 

We need to prove that whatever k , the image of the function is tangent to y=5.

 

A quadratic function has the general form of : ax^2 + bx + c.

In our case :

a=1

b=k

c=0

 

a>0 , which means the parabola opens upward.

 

Let's calculate the discriminant. I'll replace delta with D.

D = b^2 - 4ac = k^2

k^2 is always bigger or equal to 0, this means we will always have 2 distinct and real solutions to our equation (x^2 + kx=0). The solutions will obviously intersect the line of y=0.

 

Somewhere between those solutions, we can find the vertex of the function, the point where the function "turns around".

For our parabola to be tangent to the line y=5, only one point needs to intersect that line. This means that the vertex of the function must be found somewhere on the y=5 line.

We have a formula for the positions of the vertex:

So, our problem becomes proving that -D/4a =5

D=k^2

a=1

-k^2/4=5

-1/4 * k^2 - 5 = 0

This is another quadratic function with a = -1/4 , b = 0, c = -5

We calculate the discriminant :

D = b^2 - 4ac = - 4 * (-5) * (-1/4) =... < 0

We can see that the discriminant is smaller than zero (there's three negative numbers multiplied, resulting in a negative number). This means that the equation has no real solutions, thus proving our initial theory wrong.

 

The function x^2 - kx can't be tangent to the line y=5.

Thank you.

[Nineshadow]

Thank you.

Quadratic functions are so simple and quite nice.

You just need to learn some of their proprieties and you're golden.

 

(also some exercise )

[RazerZ]

Quadratic functions are so simple and quite nice.

You just need to learn some of their proprieties and you're golden.

 

(also some exercise )

I don't understand your work. You made something easy complicated and your answer is wrong as well.  Sorry, I misread your work. It's right, just another way to solve it. I did not see " Let's calculate the discriminant. I'll replace delta with D.

D = b^2 - 4ac = k^2

k^2 is always bigger or equal to 0, this means we will always have 2 distinct and real solutions to our equation (x^2 + kx=0). The solutions will obviously intersect the line of y=0."

 

Anyway, here's a gif that shows the graph with the coefficient k increasing and decreasing with the said intercepts:

 

[Arty]

I don't understand your work. You made something easy complicated and your answer is wrong as well.

 

I can prove it with this gif since you said there is no real solutions when there is actually two:

 

I don't understand your work. You made something easy complicated and your answer is wrong as well.

 

I can prove it with this gif since you said there is no real solutions when there is actually two:

 

Do you know what tangent means ?

[RazerZ]

Do you know what tangent means ?

Why do you ask?

[Arty]

Why do you ask?

the questions asks

"explain why there is no value of k for which the graph of y=x^2+kx is tangent to the line y=5"

[RazerZ]

the questions asks

"explain why there is no value of k for which the graph of y=x^2+kx is tangent to the line y=5"

Do you know what secant is?