https://file.aiquickdraw.com/mathbot/file/1726157853843h81mjnbo.jpg

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.

Să se determine funcția de gradul doi f : ℝ → ℝ, f(x) = ax² + bx + c, dacă: graficul trece prin punctele A(0, -1), B(3, -1) și este tangent dreptei y = 2.

Solve a quadratic equation problem where the function passes through the points A(0, -1) and B(3, -1) and is tangent to the line y = 2. Explore concepts of quadratic equations, derivatives, and tangency conditions. Suitable for Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation