If P(-3,-33) is a point on f(x) find the equation of the secant line PQ. Given Q is a point on f(x) when x=4. f(x)=2x-3x^2

Answer 1

Solution:

find point Q coordinates x=4, y(x)=f(x)=2x-3x² , y(4) = 2(4) -3(4)² = 8-3(16) = 8- 48 = -40 and point Q is

Q ( 4,-40), and point P(-3,-33)

The equation of the secant line is that of the line that passes through these points

First find the slope m = y2-y1/x2-x1 = -33 -(-40) / (-3-4) = (-33 + 40) / (-3-4) = 7 / -7 = -1

The slope intercept form of the equation is y = mx+ b , y = -1(x) + b, y= -x + b find b by substitution of any point into the equation, lets substitute point Q ( 4,-40) , -40 = -4 + b, b = -40+4 = -36

and the equation of the secant line PQ is y = - x - 36

using slope equation form and any point say point P ( -3,-33), m = -1 = y-(-33) / x-(-3)

-1 = (y+33) /( x+3). cross multiply , -x-3 = y +33, y = -x-3-33 = -x -36 same equation as above.

Tip: It is always a good idea to solve in two different methods to check your answer