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You need to limit the x range of the parabola to the points where the parabola crosses the red line.
Since the parabola intersects the x axis at +/- 6, and since the coefficient in x^2 is negative (the parabola is inverted) the formula for the parabola is
` ` `y = -a(x-6)(x+6)`
`= -a(x^2 - 36) = 36a - ax^2`
Also, since the intercept is at `y=8`, this implies that `a = 8/36 = 2/9` . The formula is then
`y = 8 - 2/9x^2`
To find where the parabola crosses the red line, solve
`8 - 2/9x^2 = 1`
ie, `2/9x^2 = 7`
This implies that the parabola crosses the line when `x = pm sqrt(63/2) = pm 5.6125`
Limit the x range of the parabola to between -5.6125 and +5.6125 in the RANGE field.
Whatever your calculator is , you have to calcualte the value of x for parabola intersecates red line.
I.e your parabola is `y=8-2/9x^2`
that has zero for `x=+-6`
To provivde end of graphic no over the interestion with red line `y=1` you have to find solution of equation:
`8-2/9x^2=1` `rArr` `-2/9x^2=-7` `rArr x= +-3/2 sqrt(14)`
So if you set this range claculator does not draw graphic once intersects straight line. As in figure bellow:
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