STC 110 (12 February)

Quadratic Functions

Overview of Quadratic Functions

  • Need to determine critical points such as y-intercepts and turning points.

  • Begin by rewriting the function in a standard form.

Simplification Procedure

  • The quadratic function can be expressed as: y=6x2+13x+5y = 6x^2 + 13x + 5

    • Coefficients:

    • Leading coefficient (a): 6

    • Linear coefficient (b): 13

    • Constant term (c): 5

Critical Points in Quadratics

  • The critical points include:

    • Y-intercept

    • X-intercepts

    • Turning point

Finding Y-Intercept

  • To find the y-intercept, set x=0x = 0 in the equation: y=6(0)2+13(0)+5=5y = 6(0)^2 + 13(0) + 5 = 5

    • Conclusion: The y-intercept is (0, 5).

Finding X-Intercepts

  • To find the x-intercepts, determine when y=0y = 0: 6x2+13x+5=06x^2 + 13x + 5 = 0

    • Need to factor the quadratic if possible.

    • Factors of 6: (6, 1), (2, 3)

    • Factors that result in 5: (1, 5)

Factorization Methodology

  • Use the method of trial and error, practice, and familiarize yourself with common factors.

  • Choosing the right factors can be nuanced; practice yields proficiency.

    • Example: For (6)(5)=30(6)(5) = 30, valid pairs of factors will lead to correct intercept values.

Quadratic Formula as an Alternative

  • If factorization is not straightforward, use the quadratic formula: x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

    • With the coefficients from earlier:

    • −b=−13-b = -13

    • Calculate: b2−4ac=132−4(6)(5)b^2 - 4ac = 13^2 - 4(6)(5)

    • b2=169b^2 = 169

    • 4ac=1204ac = 120

    • b2−4ac=169−120=49b^2 - 4ac = 169 - 120 = 49

    • Therefore:
      x=−13±492(6)x = \frac{-13 \pm \sqrt{49}}{2(6)}
      x=−13±712x = \frac{-13 \pm 7}{12}

    • Solutions derived will provide x-intercepts.

Completing the Analysis

  • The turning point represents the vertex of the parabola. To find the x-coordinate of the turning point (vertex), apply: x=−b2a=−1312x = \frac{-b}{2a} = \frac{-13}{12}

    • Compute the corresponding y-value by substituting back into the quadratic function to get coordinate form.

Understanding Parabola Shape

  • Determining the behavior of the parabola based on the coefficient of the quadratic term (c):

    • If c > 0:

    • Parabola opens upwards; vertex is a minimum;

    • This implies finite minimum with infinite maximum at the ends (arrows point upward).

    • If c < 0:

    • Parabola opens downwards; vertex is a maximum;

    • This indicates finite maximum with infinite minimum at the ends (arrows point downward).

    • If c=0c = 0:

    • Yields a linear expression: the graph is a line rather than a parabola.

Zero Discriminant Scenario

  • If the discriminant (part below the square root) equals zero, there is exactly one x-intercept (the vertex is on the x-axis).

Conclusion

  • Familiarize with all parts of the quadratic equation and be able to identify critical points.

  • Practice factorization and apply formulas efficiently to solve quadratic equations.

  • Always verify calculated values to ensure accuracy in solutions.

Homework Assignment

  • Solve the quadratic equation $y = 1 + 2x$ to reinforce understanding and application of concepts covered.