Sure! A quadratic function in standard form is typically written as:
\[ f(x) = ax^2 + bx + c \]
where:
- \( a \), \( b \), and \( c \) are constants,
- \( a \neq 0 \) (if \( a = 0 \), the function is linear, not quadratic).
Here are a few quick checks and properties related to quadratic functions in standard form:
1. Identifying \( a \), \( b \), and \( c \)
For the quadratic function \( f(x) = 2x^2 - 4x + 3 \):
- \( a = 2 \)
- \( b = -4 \)
- \( c = 3 \)
2. Determining the direction of the parabola
- If \( a > 0 \), the parabola opens upwards.
- If \( a < 0 \), the parabola opens downwards.
3. Finding the vertex
The x-coordinate of the vertex can be found using the formula: \[ x = -\frac{b}{2a} \]
For the example \( f(x) = 2x^2 - 4x + 3 \):
- \( x = -\frac{-4}{2(2)} = \frac{4}{4} = 1 \)
4. Finding the axis of symmetry
The axis of symmetry is the line \( x = -\frac{b}{2a} \).
5. Finding the y-intercept
Substituting \( x = 0 \) into the equation gives us the y-intercept: \[ f(0) = c \]
For \( f(x) = 2x^2 - 4x + 3 \):
- The y-intercept is \( f(0) = 3 \).
6. Solutions (Roots) of the Quadratic Equation
To find the solutions, we can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
7. Example Problem
Let's say we want to analyze the function: \[ f(x) = -3x^2 + 6x - 2 \]
-
Identify \( a \), \( b \), and \( c \):
- \( a = -3 \)
- \( b = 6 \)
- \( c = -2 \)
-
Direction of the parabola: Since \( a < 0 \), it opens downwards.
-
Vertex:
- \( x = -\frac{6}{2(-3)} = 1 \)
- Find \( f(1) = -3(1)^2 + 6(1) - 2 = 1 \)
- Vertex is at \( (1, 1) \).
-
Axis of symmetry: \( x = 1 \).
-
Y-intercept: \( f(0) = -2 \).
-
Roots (if needed):
- Use quadratic formula:
- \( b^2 - 4ac = 6^2 - 4(-3)(-2) = 36 - 24 = 12 \) (positive, two distinct roots).
- Roots will be \( x = \frac{-6 \pm \sqrt{12}}{-6} = \frac{6 \pm 2\sqrt{3}}{6} = 1 \pm \frac{\sqrt{3}}{3} \).
Feel free to ask if you have specific questions or examples!