To determine the vertex and axis of symmetry of the quadratic function \( y = x^2 - 10x + 3 \), we can use the vertex formula and the properties of a quadratic equation.
The standard form of a quadratic equation is \( y = ax^2 + bx + c \). For the given function:
- \( a = 1 \)
- \( b = -10 \)
- \( c = 3 \)
Finding the Vertex
The x-coordinate of the vertex can be found using the formula:
\[ x = -\frac{b}{2a} \]
Substituting in the values of \( a \) and \( b \):
\[ x = -\frac{-10}{2 \cdot 1} = \frac{10}{2} = 5 \]
Next, we find the y-coordinate of the vertex by substituting \( x = 5 \) back into the original equation:
\[ y = (5)^2 - 10(5) + 3 \] \[ y = 25 - 50 + 3 \] \[ y = -22 \]
Thus, the vertex of the parabola is at the point \( (5, -22) \).
Axis of Symmetry
The axis of symmetry of the parabola is a vertical line that passes through the vertex. Since the x-coordinate of the vertex is 5, the equation of the axis of symmetry is:
\[ x = 5 \]
Summary
- Vertex: \( (5, -22) \)
- Axis of Symmetry: \( x = 5 \)