Since the quadratic equation is given in standard form, we can find the vertex and roots using the vertex and quadratic formula.
Standard form is #ax^2+bx+c#. In this case:
#a=2#
#b=8#
#c=5#
The vertex formula for the #x#-coordinate is #-\frac{b}{2a}#
Plugging in yields:
#-\frac{8}{2\cdot 2}#
#=-2#
To find the #y#-coordinate, plug in the #x#-coordinate (#-2#) into the original quadratic equation, in place of #x#:
#2x^2+8x+5#
#\Rightarrow 2(-2)^2+8(-2)+5#
#\Rightarrow 2(4)-16+5#
#\Rightarrow 8-16+5#
#\Rightarrow -3#
Therefore, the vertex is #(-2,-3)#
Now, to find the #x#-intercepts, plug the corresponding #a#, #b#, and #c# values into the quadratic formula
#x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}#
#x=\frac{-8\pm\sqrt{8^2-4(2)(5)}}{2(2)}#
#x=\frac{-8\pm\sqrt{64-40}}{4}#
#x=\frac{-8\pm\sqrt{24}}{4}#
#x=\frac{-8\pm 2\sqrt{6}}{4}#
Splitting #x# up into the plus and minus values:
#x=\frac{-8+2\sqrt{6}}{4},\qquad\qquad\qquad x=\frac{-8-2\sqrt{6}}{4}#
#x=-3.2247448713916,\qquad\qquad\qquadx=-0.77525512860841#
Those are the #x#intercepts.
The #y#-intercept is simply #(0,c)#, or #(0,5)# in this case.
Bada-bing, Bada-boom.
