Question 1041928
{{{y=x^2-2x-3}}}
(a)
The x-intercepts occur where {{{ y = 0 }}}, so
{{{ x^2 - 2x - 3 = 0 }}}
{{{ ( x - 3 )*( x + 1 ) = 0 }}} 
( with practice, you can factor just by examination )
{{{ x = 3 }}}
{{{ x = -1 }}}
So, the x-intercepts are at 
( 3,0 )
( -1,0 ) 
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(b)
The y-intercepts occur where {{{ x = 0 }}}, so
{{{ y = 0^2 - 2*0 - 3 }}}
{{{ y = -3 }}}
There is one y-intercept at
(0,-3)
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(c)
The formula for the x-value of the vertex is:
{{{ x[v] = -b/(2a) }}}
{{{ x[v] = -(-2) / (2*1) }}}
{{{ x[v] = 1 }}}
Plug this value back into equation
{{{ y[v] = 1^2 - 2*1 -3 }}}
{{{ y[v] = 1 - 2 - 3 }}}
{{{ y[v] = -4 }}}
The vertex is at
( 1,-4 )
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(d)
The plot is:
{{{ graph( 400, 400, -6, 6, -6, 10, x^2 - 2x - 3 ) }}}