Find the \( x \)-intercept(s). and the coordinates of the vertex for the parabola \( y=x^{2}-6 x+5 \). If there is more than one \( x \)-interce, with commas. Fintercept(s) and the vertex of a parabola vertex:

To find the \( x \)-intercepts and the vertex of the parabola given by the equation \( y = x^2 - 6x + 5 \), follow these steps: ### 1. Finding the \( x \)-Intercepts The \( x \)-intercepts occur where \( y = 0 \). Set the equation equal to zero and solve for \( x \): \[ 0 = x^2 - 6x + 5 \] **Using the quadratic formula:** \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Where \( a = 1 \), \( b = -6 \), and \( c = 5 \): \[ x = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(5)}}{2(1)} = \frac{6 \pm \sqrt{36 - 20}}{2} = \frac{6 \pm \sqrt{16}}{2} \] \[ x = \frac{6 \pm 4}{2} \] So, the solutions are: \[ x = \frac{6 + 4}{2} = 5 \quad \text{and} \quad x = \frac{6 - 4}{2} = 1 \] **\( x \)-Intercepts:** \( 1, 5 \) ### 2. Finding the Vertex The vertex of a parabola in the form \( y = ax^2 + bx + c \) can be found using the formula: \[ \text{Vertex} = \left( -\frac{b}{2a}, \, f\left( -\frac{b}{2a} \right) \right) \] Given \( a = 1 \) and \( b = -6 \): \[ x = -\frac{-6}{2(1)} = \frac{6}{2} = 3 \] Now, substitute \( x = 3 \) back into the original equation to find the \( y \)-coordinate: \[ y = (3)^2 - 6(3) + 5 = 9 - 18 + 5 = -4 \] **Vertex:** \( (3, -4) \) ### **Summary** - **\( x \)-Intercepts:** \( 1, 5 \) - **Vertex:** \( (3, -4) \)