How do you find the equation of the secant line through the points where X has the given values: #f(x)= x^2+1#; x=-5, x=5?

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Answer 1 · 2015-10-15T20:57:46

#y=26#

We have to find points on the curve (function) for the given Xs, so:

#y_1=f(x_1)=(-5)^2+1=25+1=26#
#y_2=f(x_2)=5^2+1=25+1=26#

We have points #A(-5,26)# and #B(5,26)#.
It's obvious that line #y=26# is secant line, but let us show that:

#y=kx+n#

#26=-5k+n#
#26=5k+n#

#52=2n => n=26#
#0=10k => k=0#

Finally, #y=0x+26 => y=26#