Question

Determine the vertex and axis of symmetry of y=x2−10x+3
.(1 point)
vertex: (
); axis of symmetry: x=
Remaining Attempts : 3

Answer 1

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: \( x = 5 \)