1. First we find the gradient (slope) of the curve at x=a=2.

To do this we differentiate the function: f'(x)=-2x=-4 when x=2.

So we can write the equation of the line as y=-4x+b where b is a constant and we’ve used the slope of the tangent at x=2 to find the slope of the line.

We know the line must pass through the point where x=2 and f(2)=12-4=8. So the line must pass through the point (2,8). If we put these into the equation of the tangent line we can find b:

8=-4×2+b, so b=8+8=16 and y=-4x+16. So the blanks are -4 and 16 for y=L(x).

2. x=2.1 is close to x=2 so we can approximate f(2.1) using the line equation from (1): y=-4(2.1)+16=-8.4+16=7.6.

3. The exact value of f(2.1) is 12-2.1²=12-4.41=7.59. See how close the approximation was? The error is (7.60-7.59)/7.59=0.01/7.59=0.001318 approximately. Multiply by 100 to get the percentage error and we get 0.1318% (too high).