Question

Quadratic functions in standard form quick check

Answer 1

A quadratic function in standard form is expressed as:

\[ f(x) = ax^2 + bx + c \]

where:
- \( a \), \( b \), and \( c \) are constants,
- \( a \) is not equal to zero (if \( a = 0 \), the function is linear, not quadratic).

### Key Features

1. **Opening Direction**:
- If \( a > 0 \), the graph opens upward.
- If \( a < 0 \), the graph opens downward.

2. **Vertex**:
- The vertex of the parabola can be found using the formula:
\[ x = -\frac{b}{2a} \]
- You can then substitute this value back into the function to find the corresponding \( y \)-coordinate of the vertex.

3. **Y-Intercept**:
- The y-intercept occurs at \( (0, c) \).

4. **X-Intercepts (Roots)**:
- You can find the x-intercepts by solving the equation \( ax^2 + bx + c = 0 \) using factoring, completing the square, or the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

### Quick Check Practice Problems

1. Identify the coefficients \( a \), \( b \), and \( c \) for the function \( f(x) = 3x^2 - 6x + 2 \).

2. Determine the vertex of the function \( g(x) = -2x^2 + 8x - 5 \).

3. Find the y-intercept of the function \( h(x) = x^2 + 4x + 3 \).

4. Calculate the x-intercepts of the function \( p(x) = x^2 - 5x + 6 \) using the quadratic formula.

5. Describe the direction in which the parabola opens for the function \( q(x) = -x^2 + 3x \).

### Answers

1. Coefficients: \( a = 3 \), \( b = -6 \), \( c = 2 \)

2. Vertex:
- \( x = -\frac{8}{2(-2)} = 2 \)
- \( g(2) = -2(2)^2 + 8(2) - 5 = 7 \)
- Vertex = \( (2, 7) \)

3. Y-intercept: At \( x = 0 \), \( h(0) = 0^2 + 4(0) + 3 = 3 \) (y-intercept = (0, 3))

4. X-intercepts:
- Using the quadratic formula:
\[ x = \frac{5 \pm \sqrt{(-5)^2 - 4(1)(6)}}{2(1)} = \frac{5 \pm \sqrt{25 - 24}}{2} = \frac{5 \pm 1}{2} \]
- X-intercepts are \( x = 3 \) and \( x = 2 \).

5. Direction: Since \( a = -1 \), the parabola opens downward.