Question 927731
to find the {{{y}}} and {{{x}}} intercepts of {{{y=-2x^2-8x-5}}}, set {{{x=0}}} to find {{{y}}} intercept, then set {{{y=0}}} to find {{{x}}} intercept


{{{y=-2x^2-8x-5}}} if {{{x=0}}}, we have

{{{y=-2*0^2-8*0-5}}}

{{{y=-5}}}

{{{y}}} intercept is at ({{{0}}},{{{-5}}})


{{{y=-2x^2-8x-5}}} if {{{y=0}}}, we have

{{{0=-2x^2-8x-5}}}...use quadratic formula

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

{{{x = (-(-8) +- sqrt( (-8)^2-4*(-2)*(-5) ))/(2*(-2)) }}} 

{{{x = (8 +- sqrt( 64-40 ))/-4) }}} 

{{{x = (8 +- sqrt( 24 ))/-4) }}} 

{{{x = (8 +- 4.89)/-4) }}} 

solutions:

{{{x = (8 + 4.89)/-4) }}}

{{{x = 12.89/-4 }}}

{{{x =-3.2 }}}

or

{{{x = (8 -4.89)/-4) }}}

{{{x = 3.11/-4 }}}

{{{x =-0.8 }}}




{{{x}}} intercept is at ({{{-3.2 }}},{{{0}}}) and ({{{-0.8 }}},{{{0}}})



{{{ graph( 600, 600, -5, 5, -10, 5, -2x^2-8x-5) }}}