sketching help !

[peterpan101]

 Sketch    y=-2X10^2x -3   showing equations of asymptotes and all key coordinates

[nisha]

pretty sure this is do-able on the calculators.

[Lasercookie]

Is that ?

Anyway it doesn't matter if it is or isn't, it's pretty much the same process you'd take for any graph. To sketch this, just note what the base graph would have been, and then look at each of the transformations applied to it to get the graph we want.

In this case the base graph would be something like You should be able to sketch this graph.

So it's going to be some kind of exponential graph, with a few transformations

We can see that the transformations on the graph are
- Translated down 3 units parallel to the y-axis
- Dilated along the y-axis by a factor of 2
- Reflection in the x-axis (so it's going to be flipped upside down)

- Dilated along the x-axis by a factor of 1/2

So that gives us a pretty good idea of what the graph will look like. (an upside down exponential graph that's been moved down by three units)

Before drawing those in, let's find the intercepts and asymptotes.

What do we know about It'd have a horizontal asymptote at y = 0 (this won't change after reflection either). Our graph has been translated down by three, so we have  an asymptote at y = -3

Due to that asympote there won't be any x-intercepts (it's an upside down exponential graph). We can find the y-intercept by subbing in x = 0


So we can mark that down, and just sketch in the general shape. For showing the dilations just have the graph a bit more squished or stretched than it would have been otherwise.

[charmanderp]

If you're stuck finding a couple of basic points (y where x=...-1,0, 1... and intercepts) is the way to go. And then if you can figure out the basic shape just connect them.

[peterpan101]

thanks for your help, I have another one

sketch 2log10(x-3)

[Hancock]

y=-2X10^2x -3

It's just an exponential curve.

So we know that the horizontal asymptote will occur when y = -3.
It's been reflected in the x-axis due to the -2. N
Now all we need are the x and y intercepts.

y-int occurs when x = 0.   Therefore y(0) = -2 - 3 = -5
x - int occurs when y = 0   Therefore, once you solve this equation, you will find there is no x-intercept. (you will get log_10 (-2/3), which doesn't exist in the real number line).

[Lasercookie]

thanks for your help, I have another one

sketch 2log10(x-3)

I'm going to test you here can you give us the outline of some working on how you think you'd start off and from there we can figure out at what step you're having trouble

[peterpan101]

thanks for your help, I have another one

sketch 2log10(x-3)

I'm going to test you here can you give us the outline of some working on how you think you'd start off and from there we can figure out at what step you're having trouble

ok so you would get the inverse
x=log10(y-3)^2

converte to index so

10^x= (y-3)

now im stuck !

[Jenny_2108]

thanks for your help, I have another one

sketch 2log10(x-3)

I'm going to test you here can you give us the outline of some working on how you think you'd start off and from there we can figure out at what step you're having trouble

ok so you would get the inverse
x=log10(y-3)^2

converte to index so

10^x= (y-3)

now im stuck !

The shape of log10 graph is just similar to loge, so you just need to find vertical asymptote and x-intercept