# Uneven Graph

This isn't really about whether a graph is odd or even, but I couldn't think of a better title
How does one make a graph like f(x) = |x|, but in the third quadrant the slope is different from the 1st quadrant?

gb7nash
Homework Helper
f(x) = -|x+1| + 10?

You might want to be more specific in your question. What you said doesn't really make much sense. When you want a graph like f(x) = |x|, what do you mean?

f(x) = -|x+1| + 10?

You might want to be more specific in your question. What you said doesn't really make much sense. When you want a graph like f(x) = |x|, what do you mean?

Ok, what I mean is that if you picture the point at which the function bends as an angle, I STILL want the angle to be at the origin, however I don't want the y axis to bisect the angle

Consider that $$f(x)=\left\|x\right\| \Leftrightarrow f(x)=\left\{^{x,x \geq 0}_{-x,x < 0}$$ and that the angle is bisected iff the slope is the same on both sides. So, make it different on both sides. $$f(x)=\left\{^{ax,x \geq 0}_{-bx,x < 0}$$ where $$a,b>0;a\neq b$$

gb7nash
Homework Helper
I'm not sure if the OP is looking for a closed formula, or if a piecewise formula is ok. I can't think of a closed formula.

uart
There's nothing wrong with using a piecewise function, but if you've got something against it you can always use :

$$y = \left(\frac{a+b}{2} \right) |x| + \left(\frac{a-b}{2} \right) x$$

There's nothing wrong with using a piecewise function, but if you've got something against it you can always use :

$$y = \left(\frac{a+b}{2} \right) |x| + \left(\frac{a-b}{2} \right) x$$

Ah! This was EXACTLY what I was looking for- even better! This was for a physics problem, so really I wanted only one function. Thank you!