View an example | All parts showingFind (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)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 of the secant PQ, and determine what happens to the slope as Q approaches P; that is, as h approaches 0. Secant slope =yx=(7-4(-2+h)2)-(7-4(-2)2)hSimplify the numerator. Secant slope =yx=16h-4h2hNow 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 Simplify. y=16x+23