To determine the vertex and axis of symmetry of the quadratic equation \( y = x^2 - 10x + 3 \), we can use the vertex formula, which is given by:
- The x-coordinate of the vertex \( x = -\frac{b}{2a} \)
- The corresponding y-coordinate can be found by substituting this x-value back into the equation.
From the equation \( y = x^2 - 10x + 3 \), we identify:
- \( a = 1 \)
- \( b = -10 \)
- \( c = 3 \)
Now, we calculate the x-coordinate of the vertex:
\[ x = -\frac{-10}{2 \cdot 1} = \frac{10}{2} = 5 \]
Next, we find the y-coordinate by substituting \( x = 5 \) back into the equation:
\[ y = (5)^2 - 10(5) + 3 \] \[ y = 25 - 50 + 3 \] \[ y = -22 \]
So, the vertex is \( (5, -22) \).
The axis of symmetry, which is a vertical line that passes through the vertex, is given by the equation:
\[ x = 5 \]
Putting this all together, we have:
- Vertex: \( (5, -22) \)
- Axis of symmetry: \( x = 5 \)
So your final answer is:
Vertex: (5, -22); Axis of symmetry: x = 5