Question

graph the following function by moving the green and blue dots (if necessary). y = 4x^2

Explanation:

Step1: Find the vertex

For a quadratic function of the form $y = ax^{2}+bx + c$, in $y = 4x^{2}$, $a = 4$, $b = 0$, $c = 0$. The $x$-coordinate of the vertex is $x=-\frac{b}{2a}=0$. Substituting $x = 0$ into $y = 4x^{2}$, we get $y = 0$. So the vertex is $(0,0)$.

Step2: Find other points

Let's choose some $x$-values. When $x = 1$, $y=4\times1^{2}=4$. When $x=- 1$, $y = 4\times(-1)^{2}=4$. When $x = 2$, $y=4\times2^{2}=16$. When $x=-2$, $y = 4\times(-2)^{2}=16$. Plot the vertex $(0,0)$ and the points $(1,4),(-1,4),(2,16),(-2,16)$ and draw a smooth curve passing through these points to graph the parabola $y = 4x^{2}$.

Answer:

The graph is a parabola with vertex at the origin $(0,0)$ opening upwards, passing through points such as $(1,4),(-1,4),(2,16),(-2,16)$ etc.