Melody

Well ask your questions one at a time.

I have told you that before!

Learn from the answer, try the next one yourself and then ask the next one if you need to. 

Also when ever possible, which is just about always, type your question in. If there are diagrams then post them in as pictures.

I have told you that before too.

The smallest N is 6, the rest of my answer was incorrect. 

What region shown?

You have not included the diagram.     !

There is no diagram so how do you expect anyone to help you?

No one is here to do all your homework for you.

The question that i read can be typed into the forum. It does not need to be posted as a seperate pic.

Don't be so lazy and do your own homework!

If you want to ask a single question then  type it into the forum or post the pic on the forum.

The last one is sideways so the x and y will be swapped around.

The others must be done by rearranging the equations as well.

you need them of the form

\((x-h)^2=4a(y-k)\)

Where (h,k) is the vertex and |a| is the focal length.

Please post one question at a time.

3. What is the focus of the parabola?

y=−14x2−x+3

Enter your answer in the boxes.

\(y=−14x^2−x+3\\ y-3=−14(x^2+\frac{1}{14}x)\\ \frac{-1}{14}(y-3)=x^2+\frac{1}{14}x\\ \frac{-1}{14}(y-3)+\frac{1}{784}=x^2+\frac{1}{14}x+\frac{1}{784}\\ \frac{-1}{14}(y-3-\frac{1}{56})=(x+\frac{1}{28})^2\\ (x+\frac{1}{28})^2=\frac{-1}{14}(y-3\frac{1}{56})\\ (x+\frac{1}{28})^2=4*\frac{-1}{56}(y-3\frac{1}{56})\\\)

concave down parabola

\(Vertex=(\frac{-1}{28},3\frac{1}{56})\\ Axis\; of \;symmetry:\quad x=\frac{-1}{28}\\ focal\;length=\frac{1}{56}\\ focus=(\frac{-1}{28},3\frac{1}{56}-\frac{1}{56})\\ focus=(\frac{-1}{28},3)\\ \)

Here is the graph

Here is the original question that guest is talking about.

Hi Curly :)

\(y=(sin9x)^{6x}\\ y=e^{ln(sin9x)^{6x}}\\ y=e^{6x\cdot ln(sin9x)}\\ let\;u=6x\cdot ln(sin9x)\\ \frac{du}{dx}=6\cdot ln(sin9x)+6x \cdot \frac{9cos9x}{sin9x}\\ \qquad \text{I used product rule here}\\ \qquad \text{and also used the fact that }\\ \qquad \frac{d}{dx}ln(f(x))= \frac{f'(x)}{f(x)}\\ \frac{du}{dx}=6\cdot ln(sin9x)+6x \cdot \frac{9cos9x}{sin9x}\\ \frac{du}{dx}=6\cdot ln(sin9x)+6x \cdot 9(cot9x)\\ \frac{du}{dx}=6\left[ ln(sin9x)+9x(cot9x)\right] \\ \)

\(y=e^{6x\cdot ln(sin9x)}\\ y=e^{u}\\ \frac{dy}{du}=e^u\\ \frac{dy}{du}=e^{6x\cdot ln(sin9x)}\\ \frac{dy}{du}=e^{ ln(sin9x)\cdot 6x}\\ \frac{dy}{du}=(sin9x)^{6x}\\ \)

\(\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}\\ \frac{dy}{dx}=(sin9x)^{6x}\cdot 6\left[ ln(sin9x)+9x(cot9x) \right]\\ \frac{dy}{dx}=6(sin9x)^{6x}\left[ ln(sin9x)+9x(cot9x) \right]\\\)

If you do not understand one of the steps please explain which step/s is giving you trouble.