Question 826459
I can say that this is a parabola that looks like a bowl
turned upside down. To simplify, say it is symmetric
about the origin. That means it intersects the x-axis
at ( 6,0 ) and ( -6,0 ) The T-intercept is at ( 0,6 ).
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The general form of the parabola is:
{{{ T(x) = -a*x^2 + b*x + c }}}
The formula for the vertex is:
{{{ x[v] = -b/(2a) }}}
Since the vertex is on the y-axis,
{{{ x[v] = 0 }}}
I can say:
{{{ T(x) = -a*0^2 + b*0 + c }}}
{{{ y[v] = 6 }}}
{{{ 6 = c }}}
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So far, I have:
{{{ T(x) = -a*x^2 + b*x + 6 }}}
Plug in ( 6,0 )
{{{ 0 = -a*6^2 + b*6 + 6 }}}
{{{ -36a + 6b + 6 = 0 }}}
{{{ -6a + b + 1 = 0 }}}
(1) {{{ -6a + b = -1 }}}
also
Plug in (-6,0 )
{{{ T(x) = -a*(-6)^2 + b*(-6) + 6 }}}
{{{ 0 = -36a - 6b + 6 }}}
{{{ -36a - 6b + 6 = 0 }}}
{{{ -6a - b + 1 = 0 }}}
(2) {{{ -6a - b = -1 }}}
Add (1) and (2)
(1) {{{ -6a + b = -1 }}}
(2) {{{ -6a - b = -1 }}}
{{{ -12a = -2 }}}
{{{ a = 1/6 }}}
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Plug this back in:
(1) {{{ -6*(1/6) + b = -1 }}}
(1) {{{ -1 + b = -1 }}}
(1) {{{ b = 0 }}}
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So, the equation is:
{{{ T(x) = -(1/6)*x^2 + 6 = 0 }}}
Here's the plot:
{{{ graph( 400, 400, -8, 8, -2, 8, -(1/6)*x^2 + 6 ) }}}