The problem asks for two curve tracing processes based on the quadratic functions provided. Let's work through both of them step by step:

1. $y = x^2 + 6x - 9$

Step 1: Find the key points

• Vertex: The vertex of a quadratic function $y = ax^2 + bx + c$ is given by $x = -\frac{b}{2a}$.

  • Here, $a = 1$, $b = 6$, and $c = -9$.
  • $x = -\frac{6}{2(1)} = -3$.
  • To find the y-coordinate of the vertex, substitute $x = -3$ back into the equation:
    • $y = (-3)^2 + 6(-3) - 9 = 9 - 18 - 9 = -18$.
  • Vertex: $(-3, -18)$.

• Axis of Symmetry: The axis of symmetry is $x = -3$.

• Y-Intercept: When $x = 0$, $y = (0)^2 + 6(0) - 9 = -9$.

  • Y-intercept: $(0, -9)$.

• X-Intercepts: To find the x-intercepts, solve $x^2 + 6x - 9 = 0$ using the quadratic formula:

  • $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-6 \pm \sqrt{6^2 - 4(1)(-9)}}{2(1)} = \frac{-6 \pm \sqrt{36 + 36}}{2} = \frac{-6 \pm \sqrt{72}}{2} = \frac{-6 \pm 6\sqrt{2}}{2}$.
  • Roots: $x = -3 \pm 3\sqrt{2}$, approximately $x \approx -7.24$ and $x \approx 1.24$.

Step 2: Plot key features

• The parabola opens upwards (since $a = 1 > 0$).
• The vertex is $(-3, -18)$, the axis of symmetry is $x = -3$, and the y-intercept is $(0, -9)$.
• The x-intercepts are $x \approx -7.24$ and $x \approx 1.24$.

Step 3: Sketch the curve

• Draw the curve with the vertex at $(-3, -18)$, crossing the y-axis at $(0, -9)$, and passing through the x-intercepts.

2. $y = 2x^2 - 7x + 8$

Step 1: Find the key points

• Vertex: Using the same formula, $x = -\frac{b}{2a}$.

  • Here, $a = 2$, $b = -7$, and $c = 8$.
  • $x = -\frac{-7}{2(2)} = \frac{7}{4} = 1.75$.
  • Substitute $x = 1.75$ into the equation to find the y-coordinate:
    • $y = 2(1.75)^2 - 7(1.75) + 8 = 2(3.0625) - 12.25 + 8 = 6.125 - 12.25 + 8 = 1.875$.
  • Vertex: $(1.75, 1.875)$.

• Axis of Symmetry: The axis of symmetry is $x = 1.75$.

• Y-Intercept: When $x = 0$, $y = 2(0)^2 - 7(0) + 8 = 8$.

  • Y-intercept: $(0, 8)$.

• X-Intercepts: To find the x-intercepts, solve $2x^2 - 7x + 8 = 0$ using the quadratic formula:

  • $x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(8)}}{2(2)} = \frac{7 \pm \sqrt{49 - 64}}{4} = \frac{7 \pm \sqrt{-15}}{4}$.
  • The discriminant is negative, so there are no real x-intercepts (the curve does not cross the x-axis).

Step 2: Plot key features

• The parabola opens upwards (since $a = 2 > 0$).
• The vertex is $(1.75, 1.875)$, the axis of symmetry is $x = 1.75$, and the y-intercept is $(0, 8)$.
• Since there are no real x-intercepts, the parabola does not cross the x-axis.

Step 3: Sketch the curve

• Draw the curve with the vertex at $(1.75, 1.875)$, passing through the y-axis at $(0, 8)$.

Would you like further details or have any questions?

Related Questions:

1. How does the discriminant determine the number of x-intercepts in a quadratic equation?
2. How do you convert a quadratic function into vertex form?
3. What role does the coefficient of $x^2$ play in the shape of a parabola?
4. Can a quadratic function have complex roots and still have a real graph?
5. How do changes to the constants in a quadratic equation affect its graph?

Tip:

When graphing parabolas, always start with the vertex, as it gives the critical point from which the rest of the curve is symmetrical.