ok, you're leaving some things out.
we want to prove that:
this means that for ANY ε > 0, we can find SOME δ > 0 so that:
|x - p| < δ will guarantee that |ax2+bx+c - (ap2+bp+c)| < ε.
first, let's see what we can do in terms of re-arranging |ax2+bx+c - (ap2+bp+c)|:
|ax2+bx+c - (ap2+bp+c)| = |a(x2 - p2) + b(x - p)|
= |(x - p)||a(x + p) + b|
now, suppose we could show that: |a(x + p) + b| ≤ M, for some constant M.
then picking δ = ε/M would give:
|ax2+bx+c - (ap2+bp+c)| = |(x - p)||a(x + p) + b| < (ε/M)(M) = ε.
now if one δ works, replacing it by an even SMALLER δ will work, too. so we can set an upper limit on how large we'll let δ be.
so suppose we insist that δ ≤ 1.
what does this mean for our x's?
|x - p| < 1 means:
-1 < x - p < 1, so p-1 < x < p+1.
suppose p ≥ 0. then -(|p| + 1) = -p-1 < p-1 < x < p+1 = |p| + 1, so in this case, |x| < |p| + 1.
on the other hand, suppose p < 0.
then -(|p| + 1) = p-1 < x < p+1 < -p+1 = |p| + 1, so in this case as well, |x| < |p| + 1.
and THIS means that when |x - p| < 1,
|a(x + p) + b| = |ax + ap + b| ≤ |ax| + |ap| + |b| = |a||x| + |a||p| + |b| < |a|(|p| + 1) + |a||p| + |b| = 2|a||p| + |a| + |b|
so that is what we choose for M (so that it serves as an upper bound for the |a(x + p) + b| factor of |f(x) - f(p)|).
so, choosing δ = min(1,ε/(2|a||p|+|a|+|b|)) will definitely work.