Sketching Graphs of Real Polynomials | IB数学：实系数多项式的图像画法

📚 Sketching Graphs of Real Polynomials | IB数学：实系数多项式的图像画法

Before you sketch any polynomial graph, you must understand that a polynomial is a continuous, smooth function defined for all real numbers. The shape of its graph is determined by its degree, leading coefficient, roots, and turning points. In this article, we use a concrete example to develop a clear, step-by-step method.

在绘制任何多项式图像之前，你必须理解：多项式是定义在全体实数上的连续、光滑函数。其图像的形状由次数、首项系数、根以及转向点共同决定。在本文中，我们通过一个具体例子来建立一个清晰、分步的绘图方法。

1. The Example Polynomial | 示例多项式

Consider the real polynomial \(f(x) = 2x^3 – 3x^2 – 12x + 5\). This is a cubic function, meaning its degree is 3. Because the degree is odd and the leading coefficient is positive, we already know the end behaviour: as \(x \to +\infty\), \(f(x) \to +\infty\), and as \(x \to -\infty\), \(f(x) \to -\infty\).

考虑实系数多项式 \(f(x) = 2x^3 – 3x^2 – 12x + 5\)。这是一个三次函数，即次数为 3。由于次数为奇数且首项系数为正，我们已知其端行为：当 \(x \to +\infty\) 时，\(f(x) \to +\infty\)；当 \(x \to -\infty\) 时，\(f(x) \to -\infty\)。

2. End Behaviour | 端行为分析

For any polynomial of degree \(n\), the end behaviour is controlled by the term with the highest power. Here the leading term is \(2x^3\). Since the exponent is odd, the left and right ends go in opposite directions. Since the coefficient 2 is positive, the right end rises and the left end falls.

对于任何 \(n\) 次多项式，端行为由最高次项决定。这里首项为 \(2x^3\)。由于指数是奇数，左右两端方向相反；由于系数 2 为正，右端上升，左端下降。

As \(x \to +\infty\), \(f(x) \to +\infty\) and as \(x \to -\infty\), \(f(x) \to -\infty\)

当 \(x \to +\infty\) 时，\(f(x) \to +\infty\)；当 \(x \to -\infty\) 时，\(f(x) \to -\infty\)

3. The y-Intercept | 与 y 轴的交点

The y-intercept is simply the value of the function when \(x = 0\). For \(f(x) = 2x^3 – 3x^2 – 12x + 5\), substituting \(x = 0\) gives \(f(0) = 5\). Therefore the graph passes through the point \((0, 5)\).

y 轴截距就是函数在 \(x = 0\) 时的值。对于 \(f(x) = 2x^3 – 3x^2 – 12x + 5\)，代入 \(x = 0\) 得 \(f(0) = 5\)。因此图像经过点 \((0, 5)\)。

\(y\)-intercept: \((0, 5)\) | 与 y 轴交点：\((0, 5)\)

4. Finding the x-Intercepts | 求与 x 轴的交点

To find where the graph crosses the x-axis, we set \(f(x) = 0\). A cubic may have one, two, or three real roots. We test possible rational roots using the factor theorem. Possible rational roots are factors of 5 divided by factors of 2: \(\pm 1, \pm 5, \pm \frac{1}{2}, \pm \frac{5}{2}\).

为了求图像与 x 轴的交点，我们令 \(f(x) = 0\)。三次方程可能有一个、两个或三个实根。我们用因式定理测试可能的有理根。可能的有理根为 5 的因子除以 2 的因子：\(\pm 1, \pm 5, \pm \frac{1}{2}, \pm \frac{5}{2}\)。

Test \(x = 1\): \(f(1) = 2 – 3 – 12 + 5 = -8 \neq 0\). Test \(x = -1\): \(f(-1) = -2 – 3 + 12 + 5 = 12 \neq 0\). Test \(x = \frac{1}{2}\): \(f(\frac{1}{2}) = 2(\frac{1}{8}) – 3(\frac{1}{4}) – 12(\frac{1}{2}) + 5 = \frac{1}{4} – \frac{3}{4} – 6 + 5 = -1.5 \neq 0\). Test \(x = -\frac{1}{2}\): \(f(-\frac{1}{2}) = 2(-\frac{1}{8}) – 3(\frac{1}{4}) – 12(-\frac{1}{2}) + 5 = -\frac{1}{4} – \frac{3}{4} + 6 + 5 = 10 \neq 0\). Test \(x = 5\): \(f(5) = 250 – 75 – 60 + 5 = 120 \neq 0\). Test \(x = -5\): \(f(-5) = -250 – 75 + 60 + 5 = -260 \neq 0\). Test \(x = \frac{5}{2}\): \(f(\frac{5}{2}) = 2(\frac{125}{8}) – 3(\frac{25}{4}) – 12(\frac{5}{2}) + 5 = \frac{250}{8} – \frac{75}{4} – 30 + 5 = 31.25 – 18.75 – 30 + 5 = -12.5 \neq 0\). Test \(x = -\frac{5}{2}\): \(f(-\frac{5}{2}) = 2(-\frac{125}{8}) – 3(\frac{25}{4}) – 12(-\frac{5}{2}) + 5 = -31.25 – 18.75 + 30 + 5 = -15 \neq 0\).

测试 \(x = 1\)：\(f(1) = 2 – 3 – 12 + 5 = -8 \neq 0\)。测试 \(x = -1\)：\(f(-1) = -2 – 3 + 12 + 5 = 12 \neq 0\)。测试 \(x = \frac{1}{2}\)：\(f(\frac{1}{2}) = 2(\frac{1}{8}) – 3(\frac{1}{4}) – 12(\frac{1}{2}) + 5 = \frac{1}{4} – \frac{3}{4} – 6 + 5 = -1.5 \neq 0\)。测试 \(x = -\frac{1}{2}\)：\(f(-\frac{1}{2}) = 2(-\frac{1}{8}) – 3(\frac{1}{4}) – 12(-\frac{1}{2}) + 5 = -\frac{1}{4} – \frac{3}{4} + 6 + 5 = 10 \neq 0\)。测试 \(x = 5\)：\(f(5) = 250 – 75 – 60 + 5 = 120 \neq 0\)。测试 \(x = -5\)：\(f(-5) = -250 – 75 + 60 + 5 = -260 \neq 0\)。测试 \(x = \frac{5}{2}\)：\(f(\frac{5}{2}) = 2(\frac{125}{8}) – 3(\frac{25}{4}) – 12(\frac{5}{2}) + 5 = \frac{250}{8} – \frac{75}{4} – 30 + 5 = 31.25 – 18.75 – 30 + 5 = -12.5 \neq 0\)。测试 \(x = -\frac{5}{2}\)：\(f(-\frac{5}{2}) = 2(-\frac{125}{8}) – 3(\frac{25}{4}) – 12(-\frac{5}{2}) + 5 = -31.25 – 18.75 + 30 + 5 = -15 \neq 0\)。

None of these rational numbers is a root. This means the polynomial may have irrational or complex roots, or possibly three real irrational roots. We need to use calculus and the discriminant to understand the shape precisely.

以上有理数都不是根。这意味着该多项式可能具有无理根或复数根，也可能有三个实的无理根。我们需要使用微积分和判别式来精确理解图像形状。

5. Critical Points: Differentiation | 关键点：求导

To find turning points, we differentiate. The derivative of \(f\) is \(f'(x) = 6x^2 – 6x – 12\). Setting \(f'(x) = 0\) gives \(6(x^2 – x – 2) = 0\), which factors as \(6(x – 2)(x + 1) = 0\). Thus the critical points occur at \(x = 2\) and \(x = -1\).

为求转向点，我们求导。\(f\) 的导数为 \(f'(x) = 6x^2 – 6x – 12\)。令 \(f'(x) = 0\) 得 \(6(x^2 – x – 2) = 0\)，分解为 \(6(x – 2)(x + 1) = 0\)。因此关键点出现在 \(x = 2\) 和 \(x = -1\) 处。

\(f'(x) = 6x^2 – 6x – 12 = 6(x – 2)(x + 1)\)

For a cubic with a positive leading coefficient, the derivative is positive when \(x < -1\), negative when \(-1 < x < 2\), and positive when \(x > 2\). This tells us the function is increasing, then decreasing, then increasing again.

对于首项系数为正的三次函数，当 \(x < -1\) 时导数为正，当 \(-1 < x < 2\) 时导数为负，当 \(x > 2\) 时导数为正。这告诉我们函数先递增、再递减、再递增。

6. Coordinates of Turning Points | 转向点的坐标

Substitute \(x = -1\) into the original function: \(f(-1) = 2(-1)^3 – 3(-1)^2 – 12(-1) + 5 = -2 – 3 + 12 + 5 = 12\). So the first turning point is \((-1, 12)\).

将 \(x = -1\) 代入原函数：\(f(-1) = 2(-1)^3 – 3(-1)^2 – 12(-1) + 5 = -2 – 3 + 12 + 5 = 12\)。所以第一个转向点为 \((-1, 12)\)。

Substitute \(x = 2\): \(f(2) = 2(8) – 3(4) – 12(2) + 5 = 16 – 12 – 24 + 5 = -15\). So the second turning point is \((2, -15)\).

代入 \(x = 2\)：\(f(2) = 2(8) – 3(4) – 12(2) + 5 = 16 – 12 – 24 + 5 = -15\)。所以第二个转向点为 \((2, -15)\)。

Since the function increases to \((-1, 12)\), then decreases to \((2, -15)\), then increases again, \((-1, 12)\) is a local maximum and \((2, -15)\) is a local minimum.

由于函数递增至 \((-1, 12)\)，然后递减至 \((2, -15)\)，再递增，因此 \((-1, 12)\) 是局部极大值，\((2, -15)\) 是局部极小值。

7. The Second Derivative and Point of Inflection | 二阶导数与拐点

The second derivative is \(f”(x) = 12x – 6\). Setting \(f”(x) = 0\) gives \(x = \frac{1}{2}\). The y-coordinate is \(f(\frac{1}{2}) = 2(\frac{1}{8}) – 3(\frac{1}{4}) – 12(\frac{1}{2}) + 5 = \frac{1}{4} – \frac{3}{4} – 6 + 5 = -\frac{3}{2}\). Thus the point of inflection is \((\frac{1}{2}, -\frac{3}{2})\).

二阶导数为 \(f”(x) = 12x – 6\)。令 \(f”(x) = 0\) 得 \(x = \frac{1}{2}\)。y 坐标为 \(f(\frac{1}{2}) = 2(\frac{1}{8}) – 3(\frac{1}{4}) – 12(\frac{1}{2}) + 5 = \frac{1}{4} – \frac{3}{4} – 6 + 5 = -\frac{3}{2}\)。因此拐点为 \((\frac{1}{2}, -\frac{3}{2})\)。

This point is exactly midway between the turning points in terms of x-coordinate, which is a characteristic symmetry of cubics. The concavity changes from downward to upward at this point.

该点的 x 坐标恰好位于两个转向点的中间，这是三次函数的一个对称特征。在此点，函数的凹凸性从下凹变为上凸。

8. Number of Real Roots | 实根的数量

Because the local maximum value \(f(-1) = 12\) is positive and the local minimum value \(f(2) = -15\) is negative, the curve must cross the x-axis three times: once to the left of \(-1\), once between \(-1\) and \(2\), and once to the right of \(2\). Therefore the equation \(f(x) = 0\) has three distinct real roots.

由于局部极大值 \(f(-1) = 12\) 为正，局部极小值 \(f(2) = -15\) 为负，曲线必定与 x 轴相交三次：一次在 \(-1\) 左侧，一次在 \(-1\) 和 \(2\) 之间，一次在 \(2\) 右侧。因此方程 \(f(x) = 0\) 有三个不同的实根。

We can locate these roots approximately using the intermediate value theorem. For the left root, test between \(x = -2\) and \(x = -1\): \(f(-2) = 2(-8) – 3(4) – 12(-2) + 5 = -16 – 12 + 24 + 5 = 1\). Since \(f(-2) = 1 > 0\) and \(f(-1) = 12 > 0\), both are positive, but check further left: \(f(-3) = 2(-27) – 3(9) – 12(-3) + 5 = -54 – 27 + 36 + 5 = -40\). Since \(f(-3) < 0\) and \(f(-2) > 0\), the left root lies between \(-3\) and \(-2\).

我们可以使用介值定理近似定位这些根。对于左侧的根，测试 \(x = -2\) 和 \(x = -1\) 之间：\(f(-2) = 2(-8) – 3(4) – 12(-2) + 5 = -16 – 12 + 24 + 5 = 1\)。由于 \(f(-2) = 1 > 0\)，\(f(-1) = 12 > 0\)，两者都为正，但继续向左侧检查：\(f(-3) = 2(-27) – 3(9) – 12(-3) + 5 = -54 – 27 + 36 + 5 = -40\)。由于 \(f(-3) < 0\) 且 \(f(-2) > 0\)，左侧根位于 \(-3\) 和 \(-2\) 之间。

For the middle root, check between 0 and 1: \(f(0) = 5 > 0\) and \(f(\frac{1}{2}) = -\frac{3}{2} < 0\), so the root lies between 0 and \(\frac{1}{2}\). For the right root, check between 2 and 3: \(f(2) = -15 < 0\) and \(f(3) = 2(27) - 3(9) - 12(3) + 5 = 54 - 27 - 36 + 5 = -4\). Still negative. Check \(f(4) = 2(64) - 3(16) - 12(4) + 5 = 128 - 48 - 48 + 5 = 37\). Since \(f(3) < 0\) and \(f(4) > 0\), the right root lies between 3 and 4.

对于中间的根，检查 0 和 1 之间：\(f(0) = 5 > 0\)，\(f(\frac{1}{2}) = -\frac{3}{2} < 0\)，因此根位于 0 和 \(\frac{1}{2}\) 之间。对于右侧的根，检查 2 和 3 之间：\(f(2) = -15 < 0\)，\(f(3) = 2(27) - 3(9) - 12(3) + 5 = 54 - 27 - 36 + 5 = -4\)，仍为负。检查 \(f(4) = 2(64) - 3(16) - 12(4) + 5 = 128 - 48 - 48 + 5 = 37\)。由于 \(f(3) < 0\) 且 \(f(4) > 0\)，右侧根位于 3 和 4 之间。

9. Summary Table of Key Features | 关键特征汇总表

Feature	Value	Meaning
y-intercept	\((0, 5)\)	Point where graph crosses y-axis
Local maximum	\((-1, 12)\)	Highest point before decreasing
Local minimum	\((2, -15)\)	Lowest point before increasing
Point of inflection	\((\frac{1}{2}, -\frac{3}{2})\)	Where concavity changes
Real roots	One in \((-3, -2)\), one in \((0, \frac{1}{2})\), one in \((3, 4)\)	Three distinct real roots

特征：y 轴截距为 \((0, 5)\)，局部极大值为 \((-1, 12)\)，局部极小值为 \((2, -15)\)，拐点为 \((\frac{1}{2}, -\frac{3}{2})\)，实根分别位于 \((-3, -2)\)、\((0, \frac{1}{2})\)、\((3, 4)\) 区间内。

10. Drawing the Sketch | 绘制草图

To draw the graph by hand, follow these steps:

要手绘图像，请按以下步骤进行：

Mark the y-intercept \((0, 5)\).

标出 y 轴截距 \((0, 5)\)。
Mark the local maximum at \((-1, 12)\) and the local minimum at \((2, -15)\).

标出局部极大值 \((-1, 12)\) 和局部极小值 \((2, -15)\)。
Mark the point of inflection at \((\frac{1}{2}, -\frac{3}{2})\).

标出拐点 \((\frac{1}{2}, -\frac{3}{2})\)。
Indicate the approximate locations of the three x-intercepts.

标注三个 x 轴交点的大致位置。
Draw a smooth curve starting from the bottom left, rising to the local maximum, falling through the inflection point to the local minimum, then rising to the top right.

从左下方开始画一条平滑曲线，上升至局部极大值，穿过拐点下降到局部极小值，再上升至右上方。

Remember that the curve must pass through all marked points smoothly, with no sharp corners. At the local maximum, the tangent is horizontal; at the local minimum, the tangent is also horizontal.

记住，曲线必须平滑地穿过所有标记点，不能有尖锐的拐角。在局部极大值处切线水平，在局部极小值处切线也水平。

11. Checking with Sign and Shape | 用符号和形状检查

A useful check is to compare the sign of the function on each interval. For \(x < -3\), \(f(x) < 0\); between \(-3\) and the first root, \(f(x) > 0\); between the middle root and 0, \(f(x) > 0\); between 0 and the middle root (which is between 0 and 0.5), the sign is positive then negative across the root; between the middle root and 2, \(f(x) < 0\); between 2 and the right root, \(f(x) < 0\); to the right of the right root, \(f(x) > 0\). In particular, the sign changes at each root because all roots have odd multiplicity.

一个有用的检查是比较每个区间上函数的符号。当 \(x < -3\) 时，\(f(x) < 0\)；在 \(-3\) 和第一个根之间，\(f(x) > 0\)；在中间的根和 0 之间，\(f(x) > 0\)；在 0 和中间的根（位于 0 和 0.5 之间）之间，符号在穿过根时由正变负；在中间的根和 2 之间，\(f(x) < 0\)；在 2 和右侧的根之间，\(f(x) < 0\)；在右侧的根右边，\(f(x) > 0\)。特别地，图像在每个根处都改变符号，因为所有根的重数均为奇数。

Another check is symmetry: for any cubic of the form \(ax^3 + bx^2 + cx + d\), the point of inflection is the centre of rotational symmetry of the graph. Here, the point \((\frac{1}{2}, -\frac{3}{2})\) is the midpoint of the line segment connecting the local maximum and local minimum: \((\frac{-1+2}{2}, \frac{12+(-15)}{2}) = (\frac{1}{2}, -\frac{3}{2})\). This confirms our calculus.

另一个检查是对称性：对于任何形式为 \(ax^3 + bx^2 + cx + d\) 的三次函数，拐点是图像的中心对称中心。这里，点 \((\frac{1}{2}, -\frac{3}{2})\) 是连接局部极大值和局部极小值的线段的中点：\((\frac{-1+2}{2}, \frac{12+(-15)}{2}) = (\frac{1}{2}, -\frac{3}{2})\)。这验证了我们的微积分计算。

12. General Strategy for All Polynomials | 所有多项式的一般绘图策略

The method used here extends to any real polynomial:

这里使用的方法可以推广到任何实系数多项式：

Determine the degree and leading coefficient to establish end behaviour.

确定次数和首项系数以判断端行为。
Compute the y-intercept by setting \(x = 0\).

令 \(x = 0\) 计算 y 轴截距。
Find the x-intercepts by solving \(f(x) = 0\) using factorisation, the factor theorem, or numerical methods.

通过因式分解、因式定理或数值方法求解 \(f(x) = 0\) 来找到 x 轴交点。
Differentiate to find critical points and classify them using the first or second derivative test.

求导找到关键点，并用一阶或二阶导数测试进行分类。
Differentiate twice to find points of inflection.

二次求导找到拐点。
Plot all key points and join them with a smooth curve respecting the increasing/decreasing intervals.

标出所有关键点，并用平滑曲线连接，遵循递增/递减区间。

For polynomials of degree 4 or higher, the same steps apply, but there may be more turning points and inflection points. Always check for roots with even multiplicity, where the graph touches the x-axis but does not cross it.

对于 4 次或更高次的多项式，同样的步骤适用，但可能有更多的转向点和拐点。始终检查偶重根的情况，此时图像接触 x 轴但不穿过它。

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Sketching Graphs of Real Polynomials | IB数学：实系数多项式的图像画法