Find X-Axis Intersections of y=(x+3)²+4: Quadratic Function Analysis

Find the intersection of the function

$y=(x+3)^2+4$

With the X

To determine where the parabola intersects the x-axis, we solve for x when $y = 0$ in the function $y = (x+3)^2 + 4$.

Set $y = 0$:

$0 = (x+3)^2 + 4$

Subtract 4 from both sides to isolate the square term:

$(x+3)^2 = -4$

We now assess the equation $(x+3)^2 = -4$. A square is always non-negative, meaning that no real number squared gives a negative result. Hence, there's no real value of x satisfying this equation.

Thus, the parabola has no intersection points with the x-axis.

Therefore, the correct answer is that there is no intersection.