Question 1128688
{{{ f(x) = x^2 + 25 }}}
The x-intercept(s) is where {{{ y=0 }}}, so
{{{ x^2 + 25 = 0 }}}
{{{ x^2 = -25 }}}
{{{ x = +sqrt( -25 ) }}}
{{{ x = 5i }}}
and, also,
{{{ x = -5i }}}
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There are 2 imaginary x-intercepts, and no
real x-intercepts
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Using the general form:
{{{ y = a*x^2 + b*x + c }}}
The x-value of the vertex is at:
{{{ x[v] = -b/(2a) }}} where
{{{ a = 1 }}}
{{{ b = 0 }}}
{{{ x[v] = 0/(2*1) }}}
{{{ x[v] = 0 }}}
Plug this back into equation:
{{{ y = 0^2 + 25 }}}
The vertex is at ( 0,25 )
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Here's the plot:
{{{ graph( 400, 400, -10, 10, -5, 100, x^2 + 25 ) }}}