Question 948104
The vertex form of a parabola's equation is generally expressed as: 

{{{y = a(x-h)^2+k}}} where ({{{h}}},{{{k}}}) is the vertex 

so, write your equation in vertex form:

{{{y = -x^2 -8x -14}}}


{{{y = -(x^2 +8x) -14}}}...complete square

{{{y = (-1)(x^2 +8x+_)-(-1)*_ -14}}}

{{{y = -(x^2 +8x+4^2)+1*4^2 -14}}}

{{{y = -(x+4)^2 +16 -14}}}

{{{y = -(x+4)^2+2}}}

({{{-4}}},{{{2}}}) is the vertex 

intercepts:

{{{0= -x^2 -8x -14}}}...multiply by {{{-1}}

{{{x^2+8x +14=0}}} ...use quadratic formula

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 

{{{x = (-8 +- sqrt( 8^2-4*1*14 ))/(2*1) }}} 

{{{x = (-8 +- sqrt( 64-56 ))/2 }}} 

{{{x = (-8 +- sqrt( 8 ))/2 }}}

{{{x = (-8 +- 2.8)/2 }}} 

{{{x = (-4 +- 1.4) }}} 

solutions:

{{{x = -4 + 1.4 }}}

{{{x =-2.6 }}}

or

{{{x = -4- 1.4 }}}

{{{x =-5.4 }}}

so, x-intercepts are at:
({{{-2.6}}},{{{0}}})
and
({{{-5.4}}},{{{0}}})

{{{drawing( 600, 600, -10,10, -15, 10,
circle(-4,2,.12),locate(-4,2,V(-4,2)),
circle(-2.6,0,.12),locate(-2.6,-0.5,p(-2.6,0)),
circle(-5.4,0,.12),locate(-5.4,-0.5,p(-5.4,0)),
 graph( 600, 600, -10,10, -15, 10, -x^2 -8x -14)) }}}