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The quadratic function in question is \( f(x) = -3x^2 + 6x + 1 \). Let's analyze the key properties:

### 1. Axis of Symmetry (A.O.S.)
The axis of symmetry for a quadratic function in the form \( f(x) = ax^2 + bx + c \) is given by:
\[
x = -\frac{b}{2a}
\]
For \( f(x) = -3x^2 + 6x + 1 \), \( a = -3 \) and \( b = 6 \). Plugging these values into the formula:
\[
x = -\frac{6}{2(-3)} = 1
\]
So, the **A.O.S.** is \( x = 1 \).

### 2. Vertex
The vertex is on the axis of symmetry, and its x-coordinate is \( x = 1 \). To find the y-coordinate, substitute \( x = 1 \) into the function:
\[
f(1) = -3(1)^2 + 6(1) + 1 = -3 + 6 + 1 = 4
\]
Thus, the vertex is at \( (1, 4) \).

### 3. Opens
Since the coefficient of \( x^2 \) (which is \( -3 \)) is negative, the parabola opens **downwards**.

### 4. Max/Min Value
Because the parabola opens downwards, the vertex represents the **maximum value** of the function, which is \( 4 \).

### 5. Domain
The domain of any quadratic function is always all real numbers: \( (-\infty, \infty) \).

### 6. Range
Since the function has a maximum value at the vertex, the range is \( (-\infty, 4] \).

### 7. y-Intercept
The y-intercept occurs where \( x = 0 \). Substituting \( x = 0 \) into the function:
\[
f(0) = -3(0)^2 + 6(0) + 1 = 1
\]
Thus, the y-intercept is \( (0, 1) \).

### 8. Solutions
To find the solutions (or x-intercepts), solve the quadratic equation \( -3x^2 + 6x + 1 = 0 \) using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
For \( a = -3 \), \( b = 6 \), and \( c = 1 \):
\[
x = \frac{-6 \pm \sqrt{6^2 - 4(-3)(1)}}{2(-3)} = \frac{-6 \pm \sqrt{36 + 12}}{-6} = \frac{-6 \pm \sqrt{48}}{-6}
\]
Simplifying further:
\[
x = \frac{-6 \pm 4\sqrt{3}}{-6}
\]
Breaking it down:
\[
x = 1 \mp \frac{2\sqrt{3}}{3}
\]
Thus, the solutions are approximately \( x \approx -0.1547 \) and \( x \approx 2.1547 \).

Let me know if you'd like further details on any of these points or additional explanations!

### Relative Questions:
1. What is the significance of the axis of symmetry in a quadratic function?
2. How do you determine if a parabola opens upwards or downwards?
3. Why is the domain of a quadratic function always all real numbers?
4. How does the vertex relate to the maximum or minimum value of the function?
5. What role does the discriminant play in determining the number of real solutions?

### Tip:
When solving quadratic equations, always check the sign of the leading coefficient \( a \) to know whether the parabola opens upwards or downwards.

Analyze the quadratic function f(x) = -3x^2 + 6x + 1 to find its axis of symmetry, vertex, direction it opens, maximum or minimum value, domain, range, y-intercept, and solutions.

Learn how to analyze the quadratic function f(x) = -3x^2 + 6x + 1. This problem involves finding the axis of symmetry, vertex, direction of opening, maximum value, domain, range, y-intercept, and solutions using algebraic methods. Suitable for students in Grades 9-10.

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