y = 3 x^2 + 2 x + 1

Frequently Asked Questions

• 找到顶点 \(y = 3 x^{2} + 2 x + 1\)

  

  $$\left(-\frac{1}{3},\frac{2}{3}\right)$$

• 找到对称轴 \(y = 3 x^{2} + 2 x + 1\)

  

  $$x=-\frac{1}{3}$$

• 以顶点形式重写 \(y = 3 x^{2} + 2 x + 1\)

  

  $$y=3{\left(x+\frac{1}{3}\right)}^2+\frac{2}{3}$$

• 求导数 \(y = 3 x^{2} + 2 x + 1\)

  

  $$y^{\prime }=6x+2$$

• 查找域 \(y = 3 x^{2} + 2 x + 1\)

  

  $$x\in \mathbb{R}$$

• $$\text{找到 }x\text{ 截距/零}$$ \(y = 3 x^{2} + 2 x + 1\)

  

  $$\text{No x-intercept}$$

• 找到 y 轴截距 \(y = 3 x^{2} + 2 x + 1\)

  

  $$y=1$$

• 找到临界点 \(y = 3 x^{2} + 2 x + 1\)

  

  $$x=-\frac{1}{3}$$

• 找到局部极值 \(y = 3 x^{2} + 2 x + 1\)

  

  $$\begin{array}{rl}& \text{The local minimum is}\frac{2}{3}\text{at}x=-\frac{1}{3}\end{array}$$

• 查找递增或递减区间 \(y = 3 x^{2} + 2 x + 1\)

  

  $$\begin{array}{rl}& \text{The increasing interval is}x\ge -\frac{1}{3}\\ & \text{The decreasing interval is}x\le -\frac{1}{3}\end{array}$$

• 查找范围 \(y = 3 x^{2} + 2 x + 1\)

  

  $$y\ge \frac{2}{3}$$

• 找到垂直渐近线 \(y = 3 x^{2} + 2 x + 1\)

  

  $$\text{No vertical asymptotes}$$

• 找到水平渐近线 \(y = 3 x^{2} + 2 x + 1\)

  

  $$\text{No horizontal asymptotes}$$

• 找到斜渐近线 \(y = 3 x^{2} + 2 x + 1\)

  

  $$\text{No oblique asymptotes}$$

• 确定是偶数、奇数还是两者都不是 \(y = 3 x^{2} + 2 x + 1\)

  

  $$\text{Neither even nor odd}$$

• 求驻点 \(y = 3 x^{2} + 2 x + 1\)

  

  $$\left(-\frac{1}{3},\frac{2}{3}\right)$$

• 找到拐点 \(y = 3 x^{2} + 2 x + 1\)

  

  $$\text{No inflection points}$$

• 测试关于原点的对称性 \(y = 3 x^{2} + 2 x + 1\)

  

  $$\text{Not symmetry with respect to the origin}$$

• 测试关于 x 轴的对称性 \(y = 3 x^{2} + 2 x + 1\)

  

  $$\text{Not symmetry with respect to the x-axis}$$

• 测试关于 y 轴的对称性 \(y = 3 x^{2} + 2 x + 1\)

  

  $$\text{Not symmetry with respect to the y-axis}$$

• 求抛物线的标准方程 \(y = 3 x^{2} + 2 x + 1\)

  

  $${\left(x+\frac{1}{3}\right)}^2=\frac{1}{3}\left(y-\frac{2}{3}\right)$$

• 找到抛物线的顶点 \(y = 3 x^{2} + 2 x + 1\)

  

  $$\left(-\frac{1}{3},\frac{2}{3}\right)$$

• 找到抛物线的焦点 \(y = 3 x^{2} + 2 x + 1\)

  

  $$F=\left(-\frac{1}{3},\frac{3}{4}\right)$$

• 求抛物线的准线 \(y = 3 x^{2} + 2 x + 1\)

  

  $$y=\frac{7}{12}$$

• $$\text{求解 }x$$ \(y = 3 x^{2} + 2 x + 1\)

  

  $$\begin{array}{rl}& x=\frac{\sqrt{-2+3y}-1}{3}\\ & x=-\frac{\sqrt{-2+3y}+1}{3}\end{array}$$

• 以极坐标形式重写 \(y = 3 x^{2} + 2 x + 1\)

  

  $$\begin{array}{rl}& r=\frac{\sin \left(\theta \right)-2\cos \left(\theta \right)-\sqrt{1-9{\cos }^2\left(\theta \right)-2\sin \left(2\theta \right)}}{6{\cos }^2\left(\theta \right)}\\ & r=\frac{\sin \left(\theta \right)-2\cos \left(\theta \right)+\sqrt{1-9{\cos }^2\left(\theta \right)-2\sin \left(2\theta \right)}}{6{\cos }^2\left(\theta \right)}\end{array}$$