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 \((h, k)\) of a parabola given by the equation \( y = ax^2 + bx + c \) can be found using:
- The x-coordinate of the vertex \( h = -\frac{b}{2a} \)
- The y-coordinate \( k = f(h) \)
In this case, we have:
- \( a = 1 \)
- \( b = -10 \)
- \( c = 3 \)
Step 1: Calculate the x-coordinate of the vertex \( h \)
\[ h = -\frac{-10}{2 \cdot 1} = \frac{10}{2} = 5 \]
Step 2: Calculate the y-coordinate of the vertex \( k \)
Now we need to find \( k \) by substituting \( h = 5 \) back into the function \( y \):
\[ k = f(5) = 5^2 - 10 \cdot 5 + 3 = 25 - 50 + 3 = -22 \]
Thus, the vertex is:
\[ \text{vertex} = (5, -22) \]
Step 3: Determine the axis of symmetry
The axis of symmetry for a parabola given by \( y = ax^2 + bx + c \) is the line \( x = h \). Therefore, in this case:
\[ \text{axis of symmetry} = x = 5 \]
Final Answer:
- Vertex: \( (5, -22) \)
- Axis of symmetry: \( x = 5 \)