Only need question #8! but I need to use the method in the example 3 that's given.

Slope of a Curve at a Point In Exercises 7-18, use the method in Example 3 to find (a) the slope of the curve at the given point P, and (b) an equation of the tangent line at P. 7. y = x2 - 5, P(2, - 1) 8. y = 7 - x2, P(2, 3) 9. y = x2 - 2x - 3, P(2, -3) 10. y = x2 - 4x, P(1, -3) 11. y = x, P(2, 8) 12. y = 2 - x3, P(1, 1) 13. y = x3 - 12x, P(1, -11) 14. y = x3 - 3x2 + 4, P(2, 0) 15 . y = x, P( - 2 , - 1/2 ) Secant Lines FIGURE 2.3 The tangent line to the curve at P is the line through P whose slope is the limit of the secant line slopes as Q - P from either side. EXAMPLE 3 Find the slope of the tangent line to the parabola y = x2 at the point (2, 4) by analyzing the slopes of secant lines through (2, 4). Write an equation for the tangent line to the parabola at this point. Solution We begin with a secant line through P(2, 4) and a nearby point O(2 + h, (2 + h)2). We then write an expression for the slope of the secant line PQ and investigate what happens to the slope as @ approaches P along the curve: Secant line slope = Ay (2+ h) - 22 _12+ 4h + 4 -4 Ax h h he+ 4h= h+ 4. h If h > 0, then Q lies above and to the right of P, as in Figure 2.4. If h