Question

Find zeros from vertex form y=4(x-2)^2

Answer 1

Explanation:

The vertex form of the equation of a parabola is given by the form:

$y=a(x-h)^2+k \\ \\ \\ Where: \\ \\ a \ \text{is the leading coefficient and (h,k) the vertex of the parabola:}$

The zeroes are the x-values that touches the x-axis, that is, those x-values at which $y=0$. Form the equation it's easy to know that at $x=2$ the output $y=0$ because:

$y=4(x-2)^2 \\ \\ If \ x=2: \\ \\ y=4(2-2)^2 \\ \\ y=4(0)^2 \\ \\ y=0$

So the only zero occurs at $x=2$