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Write a polynomial from x-intercepts and a point

Use intercepts for factors and a point for scale

A zero at $x = r$ gives a factor $x - r$ . A crossing has odd multiplicity; a touch-and-bounce has even multiplicity. For the smallest possible degree, use multiplicity 1 for a crossing and 2 for a touch. Multiply these factors and include a nonzero constant $a$ . Substitute a known point $(x_{0}, y_{0})$ to determine $a$ . This method assumes that the function is a polynomial and all its real intercepts are specified; a smooth curve alone does not establish that assumption.

Problem

The intercepts and point needed for each problem are stated explicitly. No original graph is needed.

1. A cubic polynomial crosses the x-axis at $x = - 4$ , touches it at $x = - 2$ , and passes through $(- 3, 1)$ . Find its equation.

2. A polynomial touches the x-axis at $x = 0$ and crosses it at $x = 1$ and $x = 4$ . Which expression matches these intercepts?

A. $x^{2} (x - 1) (x - 4)$ .

B. $x^{2} (x + 1) (x + 4)$ .

C. $x (x + 1) (x + 4)$ .

D. $x (x - 1) (x - 4)$ .

3. Find a polynomial of the smallest possible degree whose only real intercepts are a crossing at $x = - 1$ and a touch at $x = 2$ , and which passes through $(- 2, - 18)$ .

4. Find a polynomial of the smallest possible degree whose only real zeros are crossings at $x = - 3$ , $x = - 1$ , and $x = 2$ , and which passes through $(1, 16)$ .

Answer

f (x) = (x + 4) (x + 2)^{2}

. 2. A:

f (x) = x^{2} (x - 1) (x - 4)

. 3.

f (x) = \frac{9}{8} (x + 1) (x - 2)^{2}

. 4.

f (x) = - 2 (x + 3) (x + 1) (x - 2)

The least-degree condition matters. Intercepts alone do not fix the scale factor, and higher-degree polynomials may share the same crossings and touches.

Step-by-step solution

1. A cubic with one crossing and one touch

The crossing at $- 4$ supplies $x + 4$ , while the touch at $- 2$ supplies $(x + 2)^{2}$ . Since the polynomial is cubic, these multiplicities use all three degrees:

f (x) = a (x + 4) (x + 2)^{2} .

Substitute the point $(- 3, 1)$ :

1 = a (- 3 + 4) (- 3 + 2)^{2} = a .

Therefore

f (x) = (x + 4) (x + 2)^{2} = x^{3} + 8 x^{2} + 20 x + 16.

The squared factor is nonnegative and the remaining factor is positive near $- 2$ , so the graph touches the axis there without crossing it. At $- 4$ , the simple factor changes sign.

2. Match the intercepts to the answer choices

A touch at $0$ requires an even power of $x$ , while crossings at $1$ and $4$ require factors $x - 1$ and $x - 4$ with odd powers. Option A has exactly these factors:

f (x) = x^{2} (x - 1) (x - 4) .

Option B has zeros at $- 1$ and $- 4$ , rather than $1$ and $4$ . Option C has those wrong zeros and only a single factor of $x$ . Option D has the correct zero locations but crosses at $0$ instead of touching. Thus A is the only match. As a check, $f (2) = 4 (1) (- 2) = - 8$ .

3. Determine a fractional scale factor

The smallest possible degree is $1 + 2 = 3$ , so write

f (x) = a (x + 1) (x - 2)^{2} .

Using $(- 2, - 18)$ gives

- 18 = a (- 1) (- 4)^{2} = - 16 a, a = \frac{9}{8} .

Hence

f (x) = \frac{9}{8} (x + 1) (x - 2)^{2} = \frac{9}{8} (x^{3} - 3 x^{2} + 4) .

Checking the given point yields $\frac{9}{8} (- 1) (16) = - 18$ . The factor $x + 1$ causes a crossing at $- 1$ , while the square causes a touch at $2$ .

4. Three crossings with negative leading coefficient

Each crossing needs one linear factor for the smallest possible degree:

f (x) = a (x + 3) (x + 1) (x - 2) .

At the point $(1, 16)$ ,

16 = a (4) (2) (- 1) = - 8 a, a = - 2.

Thus

f (x) = - 2 (x + 3) (x + 1) (x - 2) = - 2 x^{3} - 4 x^{2} + 10 x + 12.

Substitution at $x = 1$ gives $- 2 - 4 + 10 + 12 = 16$ . All three factors have odd multiplicity, so the graph crosses at every stated zero.

Why the extra point is needed

Multiplying a polynomial by a nonzero constant preserves its zeros and their multiplicities. A known point away from the zeros fixes that constant. If the only supplied point is an intercept, it does not determine $a$ .

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