# Thread: Sketching graphs

1. ## Sketching graphs

Hi guys

I've been doing revision for yearlies and I've been having difficulties figuring out these questions:

1) Sketch the graph of the periodic function such that f(x) = x for – 1 < x ≤ 1 where the period of f(x) is 2.

2) Find the largest possible domain and range of the function f(t) = abs(t) – t.

3) If f(x) = x^2 – 6x + 5, sketch a graph of the region defined by the intersection of the inequalities f(x) + f(y) ≤ 0 and f(x) – f(y) ≥ 0.

Could you please help me out?

Thanx

2. Hello xwrathbringerx
Originally Posted by xwrathbringerx
Hi guys

I've been doing revision for yearlies and I've been having difficulties figuring out these questions:

1) Sketch the graph of the periodic function such that f(x) = x for – 1 < x ≤ 1 where the period of f(x) is 2.

2) Find the largest possible domain and range of the function f(t) = abs(t) – t.

3) If f(x) = x^2 – 6x + 5, sketch a graph of the region defined by the intersection of the inequalities f(x) + f(y) ≤ 0 and f(x) – f(y) ≥ 0.

Could you please help me out?

Thanx
The first two are very straightforward. The third - I'm still thinking about!

(1) Between -1 and 1, the graph is the line y = x; i.e. at 45 degrees to the axes between (-1,-1) and (1, 1), excluding (-1, -1), but including (1, 1). This covers a range of values of x of length 2, the period of the function. So this line segment is repeated between x = 1 and x = 3, and again, and again, ... Also to the left between x = -3 and x = -1, etc... See sketch.

(2) Assuming $t$ is real, the domain is all the real numbers.

If $t \ge 0, |t| = t \Rightarrow |t| - t = 0$

If $t < 0, |t| = -t \Rightarrow |t|-t = -2t$ which is $> 0$ for $t < 0$.

So the range is all the non-negative reals.

Grandad

3. Hmmmm perhaps I have to shade the regions f(x) ≤ - f(y) and f(x) ≥ f(y)</SPAN>

BUT I have no idea what the final image should look like (like the final areas shaded)

Could you please show me?

4. Hello xwrathbringerx
Originally Posted by xwrathbringerx
3) If f(x) = x^2 – 6x + 5, sketch a graph of the region defined by the intersection of the inequalities f(x) + f(y) ≤ 0 and f(x) – f(y) ≥ 0.

Could you please help me out?

Thanx
Sorry I've been slow in 'thinking'!

$f(x) + f(y) = x^2-6x+5+y^2-6y+5= (x-3)^2 + (y-3)^2 -8$

So $f(x) + f(y) = 0$ is a circle, centre $(3,3)$ radius $\sqrt8\, (= 2\sqrt2)$.

And $f(x) + f(y) \le 0$ represents the points on and inside this circle.

$f(x) - f(y) = (x^2 -6x)-(y^2-6y) =(x-3)^2-(y-3)^2$, which is a hyperbola, centre $(3,3)$ with asymptotes parallel to $x = \pm y$ (Do you know what this looks like?)

So $f(x) - f(y) \ge 0$ represents the points on and 'outside' this hyperbola (i.e. points on the far side of the curve from the centre).

Can you sketch these now?

Grandad