To find the points of intersection of the two graphs algebraically, we need to solve the system of equations:

1. $2y = 10x - x^2 - 2$
2. $x + y = 6$

Step 1: Express $y$ in terms of $x$ from equation [2]

From $x + y = 6$, we solve for $y$: $y = 6 - x$

Step 2: Substitute $y = 6 - x$ into equation [1]

Substituting $y = 6 - x$ into $2y = 10x - x^2 - 2$: $2(6 - x) = 10x - x^2 - 2$ Simplify: $12 - 2x = 10x - x^2 - 2$ Rearrange to form a standard quadratic equation: $x^2 - 12x + 14 = 0$

Step 3: Solve the quadratic equation

The quadratic equation is: $x^2 - 12x + 14 = 0$ Use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = -12$, and $c = 14$: $x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(1)(14)}}{2(1)}$ $x = \frac{12 \pm \sqrt{144 - 56}}{2}$ $x = \frac{12 \pm \sqrt{88}}{2}$ $x = \frac{12 \pm 2\sqrt{22}}{2}$ $x = 6 \pm \sqrt{22}$

Step 4: Find corresponding $y$-values

Using $y = 6 - x$:

1. For $x = 6 + \sqrt{22}$: $y = 6 - (6 + \sqrt{22}) = -\sqrt{22}$

2. For $x = 6 - \sqrt{22}$: $y = 6 - (6 - \sqrt{22}) = \sqrt{22}$

Final Answer:

The points of intersection are: $(6 + \sqrt{22}, -\sqrt{22}) \quad \text{and} \quad (6 - \sqrt{22}, \sqrt{22})$

Would you like further clarification or details on any step?

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Related Questions:

1. What are the properties of the parabola described by $2y = 10x - x^2 - 2$?
2. How does substitution compare to elimination when solving systems of equations?
3. Can these points of intersection be verified graphically?
4. What are the solutions to $x^2 - 12x + 14 = 0$ in decimal form?
5. How does the discriminant in the quadratic formula determine the nature of solutions?

Tip:

Always double-check substitution and simplification to avoid calculation errors in algebraic systems!