We are given a quadratic function $y = x^2 + (k+3)x + k+3$. We need to find the range of values for $k$ for which the quadratic function crosses the x-axis at two different points. This means that the quadratic equation $x^2 + (k+3)x + k+3 = 0$ has two distinct real roots.

1. Problem Description

We are given a quadratic function $y = x^2 + (k+3)x + k+3$. We need to find the range of values for $k$ for which the quadratic function crosses the x-axis at two different points. This means that the quadratic equation $x^2 + (k+3)x + k+3 = 0$ has two distinct real roots.

2. Solution Steps

For a quadratic equation $ax^2 + bx + c = 0$ to have two distinct real roots, the discriminant must be greater than zero. The discriminant is given by $D = b^2 - 4ac$.

In our case, $a=1$, $b = k+3$, and $c = k+3$. So the discriminant is:

$D = (k+3)^2 - 4(1)(k+3)$

We want $D > 0$, so we have:

$(k+3)^2 - 4(k+3) > 0$

We can factor out $(k+3)$:

$(k+3)[(k+3) - 4] > 0$

$(k+3)(k-1) > 0$

Now we need to find the values of $k$ that satisfy this inequality. We can analyze the sign of the expression $(k+3)(k-1)$ by considering the intervals determined by the roots $k=-3$ and $k=1$.

Case 1: $k < -3$. In this case, both $(k+3)$ and $(k-1)$ are negative, so their product is positive.

Case 2: $-3 < k < 1$. In this case, $(k+3)$ is positive and $(k-1)$ is negative, so their product is negative.

Case 3: $k > 1$. In this case, both $(k+3)$ and $(k-1)$ are positive, so their product is positive.

We want $(k+3)(k-1) > 0$, so we need either $k < -3$ or $k > 1$.

3. Final Answer

The range of $k$ for which the quadratic function crosses the x-axis at two different points is $k < -3$ or $k > 1$. In interval notation, this is $(-\infty, -3) \cup (1, \infty)$.