you need to find where the V shape crosses the U shape.
Yh, I know, but the bit that confuses me is.... You know if the gradient of modulus was negative where intersection happens, you put negative infront of the modulus equation, Can you do the same with the x^2 graph?
The way to do it without the graph is to realise that ∣2x+3∣ is actually just made up from two different straight lines, 2x+3 and −(2x+3), where the changeover happens whenever 2x+3=0 hence where the conditions on x come from.
This means that, by definition, ∣2x+3∣={2x+3,x≥−23−(2x+3),x<−23.
Yh, I know, but the bit that confuses me is.... You know if the gradient of modulus was negative where intersection happens, you put negative infront of the modulus equation, Can you do the same with the x^2 graph?
if the quadratic is inside a modulus then you can do that.