Find (a) the slope of the curve at the given point P, and (b) an equation of the tangent line at P.
y = 7 – 4x2 ,
P (-2,-9).
Solution:-
a) Start with a secant line through P(-2,-9) and Q(-2+h, 7-4(-2+h)2) nearby. Write an expression for the slope the secant PQ, and determine what happens to the slope as Q approaches P, that is , as h approaches 0.
Secant slope = 
Simplify the numerator.
Secant slope = 
Now simplify the entire fraction to get 16 – 4h. Notice that this expression approaches 16 as h approaches 0.
Thus , the slope of the curve at P(-2,-9) is 16.
b) Use this slope and the given point (-2,-9) to write the equation for the tangent line.
y – (-9) = 16(x – (-2)) This is the point-slope equation.
y = 16x + 32 – 9
y = 16 x + 23
The equation of the line tangent to y = 7 – 4x2 at P (-2,-9) is y = 16x + 23