To determine the vertex and the axis of symmetry of the quadratic function \( y = x^2 - 10x + 3 \), we can use the vertex formula.
The vertex form of a quadratic function \( y = ax^2 + bx + c \) allows us to find the x-coordinate of the vertex using the formula: \[ x = -\frac{b}{2a} \]
In your function, we identify:
- \( a = 1 \)
- \( b = -10 \)
- \( c = 3 \)
Now, let's calculate the x-coordinate of the vertex: \[ x = -\frac{-10}{2 \cdot 1} = \frac{10}{2} = 5 \]
Next, we will substitute \( x = 5 \) back into the quadratic equation to find the corresponding y-coordinate: \[ y = (5)^2 - 10(5) + 3 \] \[ y = 25 - 50 + 3 \] \[ y = -22 \]
Thus, the vertex of the quadratic function is: \[ \text{vertex} = (5, -22) \]
The axis of symmetry is given by the vertical line that passes through the x-coordinate of the vertex: \[ \text{axis of symmetry: } x = 5 \]
Final Results
- Vertex: \( (5, -22) \)
- Axis of symmetry: \( x = 5 \)