Find X-Intercepts of y=(x+1)²-5: Quadratic Function Analysis

To solve this problem, follow these steps:

The problem asks to find the intersection of the parabola with the x-axis given by the equation $y = (x+1)^2 - 5$. The x-intercepts occur where $y = 0$.

Let's solve the equation for $y = 0$:

$(x+1)^2 - 5 = 0$

Simplify and solve for $x$:

• Add 5 to both sides: $(x+1)^2 = 5$
• Take the square root of both sides: $x+1 = \pm \sqrt{5}$
• Solve for $x$ by subtracting 1 from both sides: $x = -1 \pm \sqrt{5}$

This gives two solutions:

• $x = -1 + \sqrt{5}$
• $x = -1 - \sqrt{5}$

Therefore, the x-intercepts are $(-1+\sqrt{5}, 0)$ and $(-1-\sqrt{5}, 0)$.

The correct answer is then the pair $(-1+\sqrt{5}, 0)$ and $(-1-\sqrt{5}, 0)$, which matches choice 3:

$(-\sqrt{5}-1,0),(\sqrt{5}-1,0)$

Therefore, the solution to the problem is $(-1+\sqrt{5}, 0)$ and $(-1-\sqrt{5}, 0)$