Question 639971
{{{ y = x^2 + 6x + 1 }}}
The equation is in the form {{{ y = a*x^2 + b*x + c }}}
First find the axis of symmetry.
That's a vertical line at
{{{ x = -b / (2a) }}} and {{{ y = 0 }}}
{{{ x = -6 / ( 2*1 ) = -3 }}}
Draw this line, then find the point on the line
where the minimum is, ( -3,y[min] )
{{{ y[min] = (-3)^2 + 6*(-3) + 1 }}}
{{{ y[min] = 9 - 18 + 1 }}}
{{{ y[min] = -8 }}}
So the minimum is at (-3,-8) on the axis of symmetry
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Now find the y-intercept at (0,y)
{{{ y = 0^2 + 6*0 + 1 }}}
{{{ y = 1 }}}
The y-intercept is at (0,1)
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You can plug any values you want into {{{x}}} and
get the value of {{{y}}}
I'll say {{{ x = -6 }}}
{{{ y = (-6)^2 + 6*(-6) + 1 }}}
{{{ y = 36 - 36 + 1 }}}
{{{ y = 1 }}}
So the point (-6,1) lies on the graph
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Here's the plot:
{{{ graph( 400, 400, -10, 3, -10, 10, x^2 + 6x + 1 ) }}}