The third problem asks to determine the quadratic function $f(x) = ax^2 + bx + c$, where:

• The graph passes through the points $A(0, -1)$ and $B(3, -1)$.
• The graph is tangent to the line $y = 2$.

Step-by-step solution:

1. Substitute the point $A(0, -1)$:

   Since the function passes through $A(0, -1)$, we can substitute into the function: $f(0) = a(0)^2 + b(0) + c = -1$ This simplifies to: $c = -1$

   So now, the function becomes: $f(x) = ax^2 + bx - 1$

2. Substitute the point $B(3, -1)$:

   Since the function passes through $B(3, -1)$, substitute $x = 3$ and $f(3) = -1$: $a(3)^2 + b(3) - 1 = -1$ Simplifying: $9a + 3b - 1 = -1$ $9a + 3b = 0$ $3a + b = 0$ This gives the equation: $b = -3a$

3. Use the tangency condition:

   The function is tangent to the line $y = 2$ at some point. Tangency means that at a certain point $x_0$, $f(x_0) = 2$ and the slope (i.e., the derivative) at $x_0$, $f'(x_0) = 0$.

   The derivative of $f(x) = ax^2 + bx - 1$ is: $f'(x) = 2ax + b$ At the tangency point $x_0$, the derivative is 0: $2ax_0 + b = 0$ Using $b = -3a$: $2ax_0 - 3a = 0$ Simplifying: $a(2x_0 - 3) = 0$ Since $a \neq 0$, this gives: $x_0 = \frac{3}{2}$

   Now substitute $x_0 = \frac{3}{2}$ into the equation $f(x_0) = 2$: $a\left( \frac{3}{2} \right)^2 + b\left( \frac{3}{2} \right) - 1 = 2$ Simplifying: $a\left( \frac{9}{4} \right) + b\left( \frac{3}{2} \right) - 1 = 2$ Substitute $b = -3a$: $a\left( \frac{9}{4} \right) - 3a\left( \frac{3}{2} \right) - 1 = 2$ Simplifying further: $\frac{9a}{4} - \frac{9a}{2} - 1 = 2$ Multiply through by 4 to eliminate fractions: $9a - 18a - 4 = 8$ Simplifying: $-9a - 4 = 8$ $-9a = 12$ $a = -\frac{4}{3}$

4. Find $b$:

   Now that we know $a = -\frac{4}{3}$, we can find $b$: $b = -3a = -3\left( -\frac{4}{3} \right) = 4$

Final quadratic function:

Now that we have $a = -\frac{4}{3}$, $b = 4$, and $c = -1$, the quadratic function is: $f(x) = -\frac{4}{3}x^2 + 4x - 1$

Would you like more details or have any questions? Here are some related questions to explore:

1. How do you verify that the graph is tangent to the line $y = 2$?
2. Can you calculate the vertex of the quadratic function and its significance in this context?
3. How does the discriminant help in understanding the number of solutions for a quadratic equation?
4. What does it mean geometrically when the derivative of a function equals zero?
5. How would the function change if the tangency point were at a different $y$-value?

Tip: When solving quadratic functions that involve tangency, always consider both the function and its derivative to fully capture the geometric conditions.