Math Problem Statement

Graph the function: y = 3x^2 - 6x - 1

Solution

The equation given in the image is:

y=3x26x1y = 3x^2 - 6x - 1

This is a quadratic equation of the form y=ax2+bx+cy = ax^2 + bx + c, where a=3a = 3, b=6b = -6, and c=1c = -1.

Steps to graph the function:

  1. Find the vertex: The vertex of a parabola y=ax2+bx+cy = ax^2 + bx + c can be found using the formula for the x-coordinate of the vertex: x=b2ax = \frac{-b}{2a} Substituting a=3a = 3 and b=6b = -6: x=(6)2(3)=66=1x = \frac{-(-6)}{2(3)} = \frac{6}{6} = 1 Now, substitute x=1x = 1 back into the equation to find the y-coordinate of the vertex: y=3(1)26(1)1=361=4y = 3(1)^2 - 6(1) - 1 = 3 - 6 - 1 = -4 The vertex is at (1,4)(1, -4).

  2. Determine the direction: Since a=3a = 3 is positive, the parabola opens upwards.

  3. Find the y-intercept: The y-intercept occurs when x=0x = 0. Substituting x=0x = 0 into the equation: y=3(0)26(0)1=1y = 3(0)^2 - 6(0) - 1 = -1 The y-intercept is (0,1)(0, -1).

  4. Find the x-intercepts (if any): The x-intercepts occur when y=0y = 0. Solve the equation 3x26x1=03x^2 - 6x - 1 = 0 using the quadratic formula: x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} Substituting a=3a = 3, b=6b = -6, and c=1c = -1: x=(6)±(6)24(3)(1)2(3)x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(3)(-1)}}{2(3)} x=6±36+126=6±486=6±6.936x = \frac{6 \pm \sqrt{36 + 12}}{6} = \frac{6 \pm \sqrt{48}}{6} = \frac{6 \pm 6.93}{6} So, the two solutions are: x1=6+6.9362.16,x2=66.9360.16x_1 = \frac{6 + 6.93}{6} \approx 2.16, \quad x_2 = \frac{6 - 6.93}{6} \approx -0.16 The x-intercepts are approximately (2.16,0)(2.16, 0) and (0.16,0)(-0.16, 0).

Plotting the graph:

  • The vertex is at (1,4)(1, -4).
  • The parabola opens upwards.
  • The y-intercept is (0,1)(0, -1).
  • The x-intercepts are approximately (2.16,0)(2.16, 0) and (0.16,0)(-0.16, 0).

Would you like a detailed graph or further explanation?

Related Questions:

  1. How do you find the axis of symmetry of a quadratic function?
  2. Can a parabola ever open downward, and how does that affect its vertex?
  3. How does the value of aa in the quadratic equation affect the shape of the graph?
  4. What are the implications of complex roots in a quadratic equation?
  5. How can the quadratic formula be derived from completing the square?

Tip: Always check the sign of aa in the quadratic equation to determine if the parabola opens upwards or downwards.

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Math Problem Analysis

Mathematical Concepts

Quadratic Equations
Parabolas
Vertex
Intercepts

Formulas

Vertex formula: x = -b / 2a
Quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a

Theorems

Quadratic formula
Symmetry of a parabola

Suitable Grade Level

Grades 9-11