Question

我试图在图表上绘制抛物线，我已经切换到折线图。

     private void calculate_Click(object sender, EventArgs e)
    {
        double numberA = Convert.ToDouble(valueA.Text);
        double numberB = Convert.ToDouble(valueB.Text);
        double numberC = Convert.ToDouble(valueC.Text);
        displayFormula();
        double answer1 = quadCalculator1(numberA, numberB, numberC);
        double answer2 = quadCalcualtor2(numberA, numberB, numberC);
        quadOutput.Text += answer1 + " OR " + answer2;
        this.chart1.Series["quadGraph"].Points.AddXY(answer1, 0);
        this.chart1.Series["quadGraph"].Points.AddXY(answer2, 0);
        this.chart1.Series["quadGraph"].Points.AddXY(0, numberC);
    }

然而输出是一条直线，我认为我需要更多的分数？

Answer 1

简而言之，是的，您确实需要更多数据点。

你所拥有的是情节拦截x和y轴的点，并且绘制这3个是一个好的开始但是图表引擎无法解释这三个点来自哪种数据集（sin，cos）例如，可以找到包含与任何二次方相同的截距的锯齿图。

如果你想绘制曲线本身的近似值，一个快速而肮脏的解决方案是采用答案1和答案2之间的区别（小心它们是等于或想象的情况）并计算一组点的y值在最低答案之下开始一定比例，而在最高答案之上的比例相同。然后，您可以简单地旋转结果并一次添加一个。

    private void calculate_Click(object sender, EventArgs e)
{
    double numberA = Convert.ToDouble(valueA.Text);
    double numberB = Convert.ToDouble(valueB.Text);
    double numberC = Convert.ToDouble(valueC.Text);
    displayFormula();
    double answer1 = quadCalculator1(numberA, numberB, numberC);
    double answer2 = quadCalcualtor2(numberA, numberB, numberC);
    quadOutput.Text += answer1 + " OR " + answer2;

    //this.chart1.Series["quadGraph"].Points.AddXY(answer1, 0);
    //this.chart1.Series["quadGraph"].Points.AddXY(answer2, 0);
    //this.chart1.Series["quadGraph"].Points.AddXY(0, numberC);

    // Do error checking here to determine validity of answers
    // and which is the highest and lowest of the pair

    int count = 20;
    double[,] data = GetPoints(numberA, numberB, numberC, answer1, answer2, count);
    for(int i = 0; i < count; i++)
    {
        this.chart1.Series["quadGraph"].Points.AddXY(data[i, 0], data[i, 1]);
    }
}

private double[,] GetPoints(double a, double b, double c, double xInterceptLow, double xInterceptHigh, int pointCount)
{
    double[,] output = new double[pointCount,2];

    double subRange = xInterceptLow - xInterceptHigh;
    double delta = (2* subRange) / pointCount;

    double xMin = xInterceptLow - (subRange/2);
    double xMax = xInterceptHigh + (subRange/2);

    for(int i = 0; i < pointCount; i++)
    {
        double x = xMin + ( i * delta);
        double ans = GetY(a, b, c, x);
        output[i, 0] = x;
        output[i, 1] = ans;
    }
    return output;
}

private double GetY(double a, double b, double c, double x)
{
    double answer = (a * a * x) + (b * x) + c;
    return answer;
}

Answer 2

感谢。我添加了一个for循环，计算坐标的20个值。

for (int i = -10; i < 10; i++)
{
    double pointX = i;
    double pointY = anyQuad(answer1, answer2, numberA, numberB, numberC, pointX);
    this.chart1.Series["quadGraph"].Points.AddXY(pointX, pointY);
}

适合任何需要它的人！

绘制二次曲线

2 个答案: