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 |ax^{2}+bx+c - (ap^{2}+bp+c)| < ε.

first, let's see what we can do in terms of re-arranging |ax^{2}+bx+c - (ap^{2}+bp+c)|:

|ax^{2}+bx+c - (ap^{2}+bp+c)| = |a(x^{2}- p^{2}) + 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:

|ax^{2}+bx+c - (ap^{2}+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.