Find the vertex, \( x \)-intercepts and \( y \)-intercepts of the parabola \( f(x)=x^{2}-6 x+8 \) Solution:

To analyze the parabola given by the function \( f(x) = x^2 - 6x + 8 \), we will find the vertex, \( x \)-intercepts, and \( y \)-intercept. ### 1. Finding the Vertex The vertex of a parabola in the form \( f(x) = ax^2 + bx + c \) can be found using the formula for the \( x \)-coordinate of the vertex: \[ x = -\frac{b}{2a} \] In our case, \( a = 1 \) and \( b = -6 \). Plugging in these values: \[ x = -\frac{-6}{2 \cdot 1} = \frac{6}{2} = 3 \] Now, we substitute \( x = 3 \) back into the function to find the \( y \)-coordinate of the vertex: \[ f(3) = (3)^2 - 6(3) + 8 = 9 - 18 + 8 = -1 \] Thus, the vertex of the parabola is: \[ \text{Vertex} = (3, -1) \] ### 2. Finding the \( x \)-intercepts The \( x \)-intercepts occur where \( f(x) = 0 \). We set the function equal to zero and solve for \( x \): \[ x^2 - 6x + 8 = 0 \] We can factor this quadratic equation: \[ (x - 2)(x - 4) = 0 \] Setting each factor to zero gives us the \( x \)-intercepts: \[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \] \[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \] Thus, the \( x \)-intercepts are: \[ \text{X-intercepts} = (2, 0) \quad \text{and} \quad (4, 0) \] ### 3. Finding the \( y \)-intercept The \( y \)-intercept occurs where \( x = 0 \). We substitute \( x = 0 \) into the function: \[ f(0) = (0)^2 - 6(0) + 8 = 8 \] Thus, the \( y \)-intercept is: \[ \text{Y-intercept} = (0, 8) \] ### Summary of Results - **Vertex**: \( (3, -1) \) - **X-intercepts**: \( (2, 0) \) and \( (4, 0) \) - **Y-intercept**: \( (0, 8) \)