Graphs and transformations (1 Viewer)

miester · Oct 11, 2011

every time i do a past hsc paper i always lose marks in the transformations section ie when you're give y=f(x) and you sketch y=f(x^2), y=[f(x)]^2 etc etc...the only ones i can get out are the absolute value ones, all the others cost me marks

can anyone suggest a way to do those type of questions? it would REALLY be appreciated!

D94 · Oct 11, 2011

So using the function squared questions as an example... you have $y=f(x)^{2}$ .

Let $u=f(x)$ therefore getting $y=u^{2}$ . Now, the $u=f(x)$ curve is the given curve (usually a diagram). The u axis is just the y axis.

Draw the $y=u^{2}$ , and now we use this graph and the $u=f(x)$ to compare critical points, eg. x-intercept/s, y-intercept/s, asymptotes.

So for example, when $x = 5$ , $u = 0$ , when $u = 0$ , $y = 0$ . Therefore we get $P(5,0)$

...when $x = 1$ , $u = 10$ , when $u = 10$ , $y = 100$ . Therefore we get $Q(1,100)$

...when $x = 2$ , $u = 4$ , when $u = 4$ , $y = 16$ . Therefore we get $R(2,16)$

... and so on. Then the curve will usually take its shape.

It can work for sine, cosine, inverse, absolute values, exponentials, logs... you name it

miester · Oct 11, 2011

thanks, that really helped

what approach do you take when sketching 1/f(x) and f(x^2)?

b3kh1t · Oct 11, 2011

There is generally just a set of rules for 1/f(x). They are

f(x)=0 then $\frac{1}{f(x)}=\infty$ (therefore when the original function crosses the x-axis it will become asymptote)

${f(x)}=\infty$ then f(x)=0 (so it will cross the x-axis however this is not certain so when it reaches the x-axis we draw an open circle)

if you draw the line y=1 then the same points at which the line intersects with the original graph, 1/f(x) will also pass through those points.

For the f(x^2) you can use the same method D94 has shown you.
So let u=x^2, now say you take the point A(2,4), x=2 therefore u=4. However the y value has not changed, the point A has just moved from (2,4) to (4,4).
Another example is point B(3,8), x=3 therefore u=9. The y value still remains unchanged, so the point B just moves from (3,8) to (9,8).
So the f(x) graph just becomes stretched when you do f(x^2).

D94 · Oct 11, 2011

miester said:
what approach do you take when sketching 1/f(x) and f(x^2)?

For $y = \frac{1}{f(x)}$ I tend to do the same process, ie. $y = \frac{1}{u}$ , as the hyperbola can be easily drawn, and letting $u = f(x)$ .

With $y=f(x^{2})$ , you can do the same process, but now, you have $y=f(u)$ as your given curve (ie. your x axis is now the u axis), and let/draw $u=x^{2}$ , where this is just a parabola.

miester · Oct 12, 2011

that's great! thank you so much for your help

Graphs and transformations (1 Viewer)

miester

Member

D94

New Member

miester

Member

b3kh1t

Member

D94

New Member

miester

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)