LinusCalcTips

[Arty]

Do you know what secant is?

yea.....

 

 

your gif showed no tangent lines 

[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"

yea.....

 

 

your gif showed no tangent lines 

Put two and two together.

[Ekpyrosis]

-snip-

I hate to nitpick, the math part is correct, but if you want to make a coherent proof, you have to word it differently.

You shouldn't write

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

This way you can't actually proof via contradiction what is being asked, but if you replace that with

"Let's assume that there exists a value for k so that the function is tangent to y=5. Let k be that value." then your proof is correct.

[RazerZ]

I hate to nitpick, the math part is correct, but if you want to make a coherent proof, you have to word it differently.

You shouldn't write

This way you can't actually proof via contradiction what is being asked, but if you replace that with

"Let's assume that there exists a value for k so that the function is tangent to y=5. Let k be that value." then your proof is correct.

How is the math part correct? Where did he get the y coordinate of the vertex is equal to - (delta/4a) from? I don't understand. It also is shown in the gif I posted above that there is two real solutions, not no real solutions.

 

@Arty, since the graph always results in a secant line, no tangent line can exist.

[Ekpyrosis]

How is the math part correct? Where did he get the y coordinate of the vertex is equal to - (delta/4a) from?

Just insert the value for the x coordinate (-b/2a) into the function y=ax^2+bx+c.

 

I don't understand. It also is shown in the gif I posted above that there is two real solutions, not no real solutions.

The gif shows that the functions y=x^2+kx and y=5 intersect twice. The solutions are the intersection points for any given value of k. The equation of which @Nineshadow said that it has no real solutions is a different one. If it had a solution, that solution would be the answer to the question: "For what values of k does the function y=x^2+kx have a vertex with a y-value of 5?"

[RazerZ]

Just insert the value for the x coordinate (-b/2a) into the function y=ax^2+bx+c.

 

The gif shows that the functions y=x^2+kx and y=5 intersect twice. The solutions are the intersection points for any given value of k. The equation of which @Nineshadow said that it has no real solutions is a different one. If it had a solution, that solution would be the answer to the question: "For what values of k does the function y=x^2+kx have a vertex with a y-value of 5?"

Oh that makes sense! Thanks for clearing that up.

 

Sorry @Nineshadow, that is definitely one way to solve it.

[Nineshadow]

I hate to nitpick, the math part is correct, but if you want to make a coherent proof, you have to word it differently.

You shouldn't write

This way you can't actually proof via contradiction what is being asked, but if you replace that with

"Let's assume that there exists a value for k so that the function is tangent to y=5. Let k be that value." then your proof is correct.

Ah yes , I know. I rushed to write it.It was like 3AM and I wanted to go to sleep.Not only that , but I'm not exactly an expert in English terms of mathematics.

[Nineshadow]

Just insert the value for the x coordinate (-b/2a) into the function y=ax^2+bx+c.

The gif shows that the functions y=x^2+kx and y=5 intersect twice. The solutions are the intersection points for any given value of k. The equation of which @Nineshadow said that it has no real solutions is a different one. If it had a solution, that solution would be the answer to the question: "For what values of k does the function y=x^2+kx have a vertex with a y-value of 5?"

Yours is probably a nicer solution in this case. I wanted to explain or give examples of useful stuff while working with quadratic functions, the effect of the sign of 'a' , discriminants, the vertex,etc. It still needed more explainations in that regard, especially some illustrations.

[Nineshadow]

Another way to solve it is with the proprieties of the function's parabola.We have the discriminant equal or greater than 0, that means we always have at least one solution. Consequently, the parabola will always intersect the y=0 line. The 'a' of the function is 1, 1>0.This means that the parabola opens upward. It's concave. It goes from high values, lower to the vertex (the lowest it goes actually), then back up. Since we always have solutions , and the parabola opens upward, it's obvious that the function can't be tangent to the y=5 line, since it will always intersect it at 2 points.

@Ekpyrosis