https://r2.erweima.ai/mathbot/image/1726052907589qlgakd7u.jpg

Here’s a breakdown and solution guide for each problem in the image:

### 1. **Find the domain and range of the function:**
   \( f(x) = x^2 - 5x + 7 \)

   - **Domain**: Since the function is a quadratic polynomial, the domain includes all real numbers.
     \[
     \text{Domain: } (-\infty, +\infty)
     \]
   - **Range**: To find the range, first find the vertex of the parabola. The vertex for a quadratic equation \( ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \).
     \[
     x = \frac{-(-5)}{2(1)} = \frac{5}{2}
     \]
     Now, substitute \( x = \frac{5}{2} \) into the equation to find \( f(x) \):
     \[
     f\left(\frac{5}{2}\right) = \left(\frac{5}{2}\right)^2 - 5\left(\frac{5}{2}\right) + 7 = \frac{25}{4} - \frac{25}{2} + 7 = \frac{-9}{4}
     \]
     So the minimum value of \( f(x) \) is \( \frac{-9}{4} \), and since the parabola opens upwards, the range is:
     \[
     \text{Range: } \left[\frac{-9}{4}, +\infty\right)
     \]

   - **Graph**: You can sketch the parabola using the vertex \( \left(\frac{5}{2}, \frac{-9}{4}\right) \) and the fact that it opens upward.

### 2. **A quadratic function crosses the x-axis at points (-5,0), (7,0), and passes through (-1,-8). Find the equation, domain, and range:**

   - The general form of the quadratic is \( f(x) = a(x + 5)(x - 7) \), since the x-intercepts are given.
   - Use the point (-1, -8) to find \( a \):
     \[
     -8 = a(-1 + 5)(-1 - 7) = a(4)(-8) \implies -8 = -32a \implies a = \frac{1}{4}
     \]
     So, the quadratic function is:
     \[
     f(x) = \frac{1}{4}(x + 5)(x - 7)
     \]

   - **Domain**: All real numbers.
     \[
     \text{Domain: } (-\infty, +\infty)
     \]

   - **Range**: The vertex is midway between \( x = -5 \) and \( x = 7 \), so the x-coordinate of the vertex is \( x = 1 \). Substitute into the equation to find the minimum value:
     \[
     f(1) = \frac{1}{4}(1 + 5)(1 - 7) = \frac{1}{4}(6)(-6) = -9
     \]
     So the range is:
     \[
     \text{Range: } [-9, +\infty)
     \]

### 3. **A quadratic function has a vertex at (-2,7) and passes through (-4,5). Find the equation, domain, and range:**

   - The vertex form of the quadratic is \( f(x) = a(x + 2)^2 + 7 \).
   - Use the point (-4, 5) to find \( a \):
     \[
     5 = a(-4 + 2)^2 + 7 \implies 5 = 4a + 7 \implies a = \frac{-1}{2}
     \]
     So the function is:
     \[
     f(x) = \frac{-1}{2}(x + 2)^2 + 7
     \]

   - **Domain**: All real numbers.
     \[
     \text{Domain: } (-\infty, +\infty)
     \]

   - **Range**: Since the parabola opens downwards, the maximum value is at the vertex \( y = 7 \), and it decreases towards negative infinity.
     \[
     \text{Range: } (-\infty, 7]
     \]

### 4. **A quadratic function passes through points (-2,1), (4,-3), and (-6,47). Find the equation, domain, and range:**

   - You need to set up a system of equations for the general quadratic form \( f(x) = ax^2 + bx + c \) using the given points:
     \[
     f(-2) = 1: 4a - 2b + c = 1
     \]
     \[
     f(4) = -3: 16a + 4b + c = -3
     \]
     \[
     f(-6) = 47: 36a - 6b + c = 47
     \]
   - Solve this system to find \( a \), \( b \), and \( c \).

   - **Domain**: All real numbers.
     \[
     \text{Domain: } (-\infty, +\infty)
     \]

   - **Range**: To find the range, you need to calculate the vertex.

### 5. **The quadratic function is \( f(x) = ax^2 - 2ax + 1 \):**

   a. **Find the values of \( a \) so that the function is always positive**:
      - For the function to be always positive, its discriminant must be less than zero. The discriminant \( \Delta \) is:
        \[
        \Delta = (-2a)^2 - 4a(1) = 4a^2 - 4a = 4a(a - 1)
        \]
        For \( \Delta < 0 \), \( 0 < a < 1 \).

   b. **Can the function ever be always negative?**:
      - No, the function cannot be always negative because the parabola opens upwards for positive \( a \).

   c. **Find \( a \) so that the function touches the x-axis**:
      - The function touches the x-axis when the discriminant is zero. Set the discriminant to zero:
        \[
        4a(a - 1) = 0 \implies a = 0 \text{ or } a = 1
        \]
      The function will touch the x-axis at \( a = 1 \).

Would you like any specific part explained in more detail? Here are 5 related questions to expand your understanding:

1. How do we find the vertex of a quadratic function using its general form?
2. What are the characteristics of a quadratic function's graph?
3. How do we determine the domain and range of more complex functions?
4. How does the discriminant determine the nature of the roots of a quadratic function?
5. What does the value of \( a \) tell us about the orientation of a parabola?

**Tip**: Always check the sign of \( a \) to determine whether a parabola opens upwards or downwards.

Find the domain, range, and solve various quadratic functions as per the given points and conditions.

Explore step-by-step solutions for solving quadratic functions, including domain and range identification, graphing, and using discriminants. Perfect for students in Grade 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation