Question 700487
to find the {{{y-intercepts}}} of {{{f(x)=x^2-2x-3}}} set {{{x=0}}} and solve for {{{y}}}

{{{f(x)=0^2-2*0x-3}}}

{{{f(x)=-3}}}

so, the {{{y-intercept}}} is at ({{{0}}},{{{-3}}})



to find the {{{x-intercepts}}} of {{{f(x)=x^2-2x-3}}} set {{{f(x)=0}}} and solve for {{{x}}}

{{{x^2-2x-3=0}}}........use quadratic formula


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

{{{x = (-(-2) +- sqrt( (-2)^2-4*1*(-3) ))/(2*1) }}}


{{{x = (2 +- sqrt(4+12 ))/2 }}}

{{{x = (2 +- sqrt(16 ))/2 }}}


{{{x = (2 +- 4)/2 }}}

solutions:

{{{x = (2 + 4)/2 }}}

{{{x = 6/2 }}}

{{{x = 3 }}}

or

{{{x = (2 - 4)/2 }}}

{{{x = -2/2 }}}

{{{x = -1 }}}

so, the {{{x-intercepts}}} are at ({{{3}}},{{{0}}}) and ({{{-1}}},{{{0}}})


{{{ graph( 600, 600, -10, 10, -10, 10, x^2-2x-3) }}}