Closed Calculus

[Budak69]

Paturo po sa number 7 at 4
Salamat po
Respect

 

[pinapit]

#7 first find the intersection of the line and the curve by substituting y=3 in y=x^2-2x and will give two value = 3 or -1. therefore there are two points of intersection point (3,3) and point (-1,3) using point (3,3), get the derivative of the curve giving dy/dx = 2x-2, since x=3, dy/dx = 4, using equation of line y=mx+b substitute the point (3,3) and m=dy/dx=4, 3=4(3)+b, b= -9, therefore the equation of tangent is y =4x-9, arrange it 4x-y=9. the slope of normal line is = to the negative reciprocal of the slope of tangent or = -1/4 and use again the eqn of line y=mx+b to get its equation, sorry medyo mahaba hehe.