∿Graphing Polynomials Practice Problems

The -intercept is . Substituting: . The curve passes through .

Every term containing vanishes at , leaving only the constant. So . For a polynomial in standard form the -intercept is always just the constant term — no arithmetic needed.

The exponent on is 2, which is even, so the curve comes down to the axis at and turns back without passing through.

appears to the first power, an odd multiplicity, so the graph passes straight through the axis at .

Add the exponents for the degree: . The limit is , so .

Substitute zero into each factor: . Note the squared bracket gives , not — squaring kills the sign.

There are two zeros: with multiplicity 1, and with multiplicity 3. Both multiplicities are odd, so the graph crosses at both — 2 crossings. The one at flattens as it crosses, but it still crosses.

Multiplicity 2 is even, so it touches. With both zeros even and the leading coefficient positive, this graph never goes below the axis at all — it sits on it twice and rises away on both sides.

Substitute the known point : . So . Without that extra point is unknowable — every vertical stretch has the same zeros.

Substitute : . So . A fractional stretch factor is perfectly normal — nothing requires to be an integer.

Trace the shape. Coming in high from the left it falls to cross at , continues down through , rises to touch the axis at and turn back down, falls again, then crosses at and climbs away. That is a turn at the bottom before , the turn at itself, and another bottom before — 3 turning points, which is the maximum for a quartic.

. Two minus signs — the one out front and the one from — cancel, so the intercept is positive.