Question: turning point of $y=4x^2$
1. **State the problem:** Find the turning point of the function $y=4x^2$.
2. **Recall the formula:** The turning point of a quadratic function $y=ax^2+bx+c$ occurs at $x=-\frac{b}{2a}$.
3. **Identify coefficients:** Here, $a=4$, $b=0$, and $c=0$.
4. **Calculate the x-coordinate of the turning point:**
$$x=-\frac{0}{2\times4}=0$$
5. **Find the y-coordinate by substituting $x=0$ into the function:**
$$y=4(0)^2=0$$
6. **Conclusion:** The turning point is at the coordinate $(0,0)$.
This point is a minimum because the coefficient $a=4$ is positive, so the parabola opens upwards.