Find Intersection Points of y = (4x+1)(x-2) with the X-Axis

Determine the points of intersection of the function

$y=(4x+1)(x-2)$

With the X

To find the points of intersection of the function $y = (4x+1)(x-2)$ with the x-axis, we need to solve this equation where $y = 0$. Therefore, we set $(4x+1)(x-2) = 0$.

By applying the Zero Product Property, we know that either $4x+1 = 0$ or $x-2 = 0$.

• Solving the first equation $4x+1 = 0$, we have:
$4x = -1 \Rightarrow x = -\frac{1}{4}$
• Solving the second equation $x-2 = 0$, we have:
$x = 2$

Therefore, the points of intersection with the x-axis, where the function equals zero, are $(-\frac{1}{4}, 0)$ and $(2, 0)$.

Thus, the solution to the problem is $(2,0),(-\frac{1}{4},0)$.