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How would you go about writing down the solution set of a quadratic relation with only one x-intercept?

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y = k(x-a)\^2 (k <> 0)

Note that if there is only one intercept, then it must be the vertex of the parabola. The location of that vertex is a (in the above formula) and k is a scale factor whose magnitude stretches or squishes the parabola and the sign of k determines whether it points up or down.
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I think you are talking about quadratic inequalities right? If it just has one intercept that means the it will always be positive or zero  or always be negative or zero. So your solution interval intervals will reflect that.

Some examples  x^2 > 0  (- &infin; , 0) U (0, &infin;)  and. x^2 &geq; 0  has (- &infin; , &infin;) , x^2 < 0  the solution set is empty.

Since  quadratics with one solution are essentially horizontal translations of this graph the form  (x &pm; k)^2  can you finish it from there?
by

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