Determine the vertex and axis of symmetry of y=x2−10x+3

.(1 point)
vertex: (
); axis of symmetry: x=
Remaining Attempts : 3

1 answer

To find the vertex and axis of symmetry of the quadratic function \( y = x^2 - 10x + 3 \), we can use the formulas for the vertex and axis of symmetry of a parabola.

The standard form of a quadratic function is \( y = ax^2 + bx + c \). In your case, \( a = 1 \), \( b = -10 \), and \( c = 3 \).

Step 1: Find the Axis of Symmetry

The formula for the axis of symmetry is given by:

\[ x = -\frac{b}{2a} \]

Substituting the values:

\[ x = -\frac{-10}{2 \cdot 1} = \frac{10}{2} = 5 \]

Step 2: Find the Vertex

To find the vertex, we need the x-coordinate, which we already calculated as \( x = 5 \). Now, substitute \( x = 5 \) back into the equation to find the y-coordinate of the vertex:

\[ y = (5)^2 - 10(5) + 3 \] \[ y = 25 - 50 + 3 \] \[ y = -22 \]

Conclusion

The vertex of the parabola is at \( (5, -22) \), and the axis of symmetry is \( x = 5 \).

Thus:

  • Vertex: \( (5, -22) \)
  • Axis of Symmetry: \( x = 5 \)