Question

What's the vertex and axis of symmetry of y=2x^2+8x+3 ?

Answer 1

well, the squared varaible is the "x", that means is a vertical parabola, so the axis of symmetry, will be x = "the x-coordinate of the vertex", whatever that coordinate happens to be, so let's check using the coefficients.


`\bf \textit{ vertex of a vertical parabola, using coefficients}\\\\ \begin{array}{llccll} y = &{{ 2}}x^2&{{ +8}}x&{{ +3}}\\ &\uparrow &\uparrow &\uparrow \\ &a&b&c \end{array}\qquad \left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right) \\\\\\ \left( -\cfrac{8}{2(2)}~~,~~ 3-\cfrac{8^2}{4(2)} \right)`