Find X-Intercepts of y=(x+1¼)²: Perfect Square Intersection

Find the intersection of the function

$y=(x+1\frac{1}{4})^2$

With the X

To find the intersection of the function $y = (x + 1\frac{1}{4})^2$ with the x-axis, we set $y = 0$ since intersections on the x-axis have a $y$-coordinate of zero.

Therefore, our equation becomes:

$(x + 1\frac{1}{4})^2 = 0$.

To solve this equation, take the square root of both sides:

$x + 1\frac{1}{4} = 0$.

Next, solve for $x$ by subtracting $1\frac{1}{4}$ from both sides:

$x = -1\frac{1}{4}$.

Thus, the intersection point of the function with the x-axis is $(-1\frac{1}{4}, 0)$.

After checking the provided answer choices, the correct choice is:

: $(-1\frac{1}{4},0)$

Therefore, the solution to the problem is $(-1\frac{1}{4}, 0)$.