Question 29731
For the first function: {{{y = 4x^2}}}, you know (or should know) that this is the equation for a parabola that opens upwards (the coefficient of x^2 is positive) and the x-coordinate of the vertex is given by: {{{x = -b/2a)}}}.
The a and b come from the standard form for the quadratic equation: {{{ax^2 + bx + c}}}, so the vertex of the parabola is: {{{x = 0}}} because b = 0, and, in your equation, when x = 0 then y = 0, so the vertex of the parabola is at (0, 0), the origin of the coordinate system.
The x-intercepts can be found by setting y = 0 and solving for x.
{{{0 = 4x^2}}} 
{{{x = 0}}} and {{{x = 0}}} or, there are no x-intercepts which makes sense when you realise that if the parabola opens upwards and the vertex is at the origin, then the parabola never crosses the x-axis.
Here's the graph:
{{{graph(300,200,-3,3,-3,5,4x^2)}}}

For the second function, you can apply a similar analysis and find that the parabola opens upwards but the vertex is at (0, -2) and the x-intercepts are at x = {{{sqrt(2)}}} and x = -{{{sqrt(2)}}}
Here's the graph:
{{{graph(300,200,-5,5,-5,5,x^2-2)}}}