Question

The graph of this quadratic function is a parabola. What is the x-coordinate of the vertex? y = x^2 - 3

Answer 1

-b/2a is the correct answer:)

Explanation:

Answer 2

`\bf y = x^2 - 3\implies y=1x^2+0x-3\qquad now \\\\\\ \textit{vertex of a parabola}\\ \\ \begin{array}{llccll} y=&1x^2&+0x&-3\\ &\uparrow &\uparrow &\uparrow \\ &a&b&c \end{array}\qquad \left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)`

so, the x-coordinate is
`\bf -\cfrac{{{ b}}}{2{{ a}}}`