How about y=x^2+2x+1Surely you can come up with an example. Think one of the simplest polynomials there is. Translate it so it's vertex appears at [MATH]x=r[/MATH], rather than [MATH]x=0[/MATH]
How about y=x^2+2x+1
P(x)=r^2+2x+1? I'm sorry this one question is just so complicated to me, forgive my brain.that only touches the x-axis at [MATH]r[/MATH] if [MATH]r=-1[/MATH]
move the parabola [MATH]y=x^2[/MATH] over so the vertex is at [MATH]x=r[/MATH]what do you get?
P(x)=r^2+2x+1? I'm sorry this one question is just so complicated to me, forgive my brain.