Question

question 18 of 23
1 point
which part of the quadratic formula finds the x value of the vertex of
a parabola?
select the correct response:
$b \pm \sqrt{b^2 + 4ac}$
$\sqrt{b^2 - 4ac}$
$-\frac{b}{2a}$
$-b \pm \sqrt{b^2 - 4ac}$

Explanation:

The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). The vertex of a parabola (for a quadratic function \( y = ax^2 + bx + c \)) lies on the axis of symmetry, and the formula for the axis of symmetry (which gives the x - value of the vertex) is \( x = -\frac{b}{2a} \). This is the part of the quadratic formula that determines the x - coordinate of the vertex (before considering the \( \pm \sqrt{b^2 - 4ac} \) part which is related to the roots). The other options: \( b\pm\sqrt{b^2 + 4ac} \) is not part of the standard quadratic formula, \( \sqrt{b^2 - 4ac} \) is the discriminant (related to the nature of roots), and \( -b\pm\sqrt{b^2 - 4ac} \) is part of the formula for finding the roots (not the vertex's x - value).

Answer:

\( -\frac{b}{2a} \) (the option with \( -\frac{b}{2a} \))