The graph of p(x) = x^{2} + 6x + 9 intersects | vedclass

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(C) To determine how the graph of $p(x) = x^{2} + 6x + 9$ intersects the $X$-axis,we find the zeros of the polynomial by setting $p(x) = 0$.$x^{2} + 6

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(C) To determine how the graph of $p(x) = x^{2} + 6x + 9$ intersects the $X$-axis,we find the zeros of the polynomial by setting $p(x) = 0$.$x^{2} + 6x + 9 = 0$This is a perfect square trinomial,which can be factored as $(x + 3)^{2} = 0$.Thus,$x = -3$ is the only root (a repeated root).Since there is only one distinct real root,the parabola represented by the quadratic polynomial touches the $X$-axis at exactly one point,which is $(-3, 0)$.

The graph of $p(x) = x^{2} + 6x + 9$ intersects the $X$-axis $\ldots \ldots \ldots \ldots$

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