Question 1056864
Let the form of the parabola =
{{{ y = a*x^2 + b*x + c }}}
The formula for the x-value of the vertex is:
{{{ x[v] = -b/(2a) }}}
The vertex is at: ( -4, 2 )
{{{ -4 = -b/(2a) }}}
{{{ 4 = b/(2a) }}}
{{{ b = 8a }}}
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The parabola goes through ( 0, -30 )
{{{ y = a*x^2 + b*x + c }}}
{{{ -30 = a*0^2 + b*0 + c }}}
{{{ c = -30 }}}
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{{{ y = a*x^2 + b*x + c }}}
{{{ y = a*x^2 + 8a*x - 30 }}}
Find {{{ a }}} using ( -4, 2 ) the vertex
{{{ 2 = a*(-4)^2 + 8a*(-4) - 30 }}}
{{{ 32 = 16a - 32a }}}
{{{ -16a = 32 }}}
{{{ a = -2 }}}
and
{{{ b = 8a }}}
{{{ b = -16 }}}
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The equation is:
{{{ y = -2x^2 - 16x - 30 }}}
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Does it go through ( -4,2 ) ?
{{{ 2 = -2*(-4)^2 - 16*(-4) - 30 }}}
{{{ 2 = -2*16 + 64 - 30 }}}
{{{ 2 = -32 -30 + 64 }}}
{{{ 2 = -62 + 64 }}}
{{{ 2 = 2 }}}
OK
Does it go through ( 0,-30 )
Yes
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Find x-intercepts ( roots )
{{{ -2x^2 - 16x - 30 = 0 }}}
{{{ x^2 + 8x + 15 = 0 }}}
{{{ ( x + 3 )*( x + 5 ) = 0 }}}
{{{ x = -3 }}}
{{{ x = -5 }}}
The x-intercepts are:
( -3, 0 )
( -5, 0 )
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Here's the plot:
{{{ graph( 400, 400, -8, 4, -10, 4, -2x^2 - 16x - 30 ) }}}