y=x^2-4 1.sketch 2.points the curve crosses x axis 3.gradient of those 2 points. keep it simple to understand , use drawings too

1. Sketch: The equation given is of a parabola, y = x^2 - 4. The graph of this equation would look like a U shape, with the vertex of the parabola at the point (0, -4). Here's a simple sketch:

10 |----------------*----------------|
   |                 |                |
8  |                 |                |
   |                 |                |
6  |                 |                |
   |                 |                |
4  |                 |                |
   |                 |                |
2  |                 |                |
   |                 |                |
0  |--------*--------*--------|
  -4       -2        0        2       4

The "*" represents the points where the curve crosses the x-axis.

2. Points the curve crosses x axis: The curve crosses the x-axis where y = 0. So, we set y = 0 in the equation and solve for x:

0 = x^2 - 4 x^2 = 4 x = ±2

So, the curve crosses the x-axis at the points (-2, 0) and (2, 0).

3. Gradient of those 2 points: The gradient of a curve at a particular point is given by the derivative of the function at that point. The derivative of y = x^2 - 4 is dy/dx = 2x.

At the point (-2, 0), the gradient is 2*(-2) = -4. At the point (2, 0), the gradient is 2*2 = 4.