How do you find the equation of the secant line of #f(x)=x^2-5x# through the points [1,8]?

2 Answers
Dec 20, 2016

The line will pass through the points #(1, f(1))# and #(8, f(8))#

#f(1) = 1^2 - 5(1)#

#f(1) = -4#

#f(8) = 8^2 - 5(8)#

#f(8) = 24#

The line will pass through the points #(1, -4)# and #(8, 24)#

The slope is #m =(24 - (-4))/(8 - 2) = 28/7 = 4#

The point-slope form of the equation of a line is:

#y = m(x - x_0) + y_0#

Substitute #4# for m, 1 for #x_0# and -4 for #y_0#

#y = 4(x - 1) - 4#

#y = 4x - 4 - 4#

#y = 4x - 8#

Dec 20, 2016

# y = 4x-8 #

Explanation:

We have #f(x) = x^2 - 5x#

When #x=1 => f(x) = 1-5 = -4#
When #x=8 => f(x) = 64-40 = 24#

So the required secant line passes through the points #(1, -4)# and #(8, 24)#.

We can calculate the slope of the secant line using

#m=(Delta y)/(Delta x) = (24-(-4)) / (8-1) = 28/7 = 4#

So using the factthat the line passes through #(1,-4)# and has slope #4# and Using the formula #y - y_0 = m(x - x_0) #, the required equation is given by:

# y - (-4)=4(x-1) #
# :. y +4 = 4x-4 #
# :. y = 4x-8 #

NB: We could have equally used the other coordinate

Which we can confirm graphically:
enter image source here