Find Equation of the Secant Line

Find the equation of the secant line that intersects the curve y = x^2 - 4 at x = -1 & x = 2.

P. S. Make up a similar problem for me to do on my own.

Thank you.

 

Find the equation of the secant line that intersects the curve y = x^2 - 4 at x = -1 & x = 2

A secant line is simply a linear equation and with two given points you can find the equation.

The two points on the secant line are:
y = (-1)^2 - 4 =-3 =>(-1,-3)

y = 2^2 - 4 =0 =>(2,0)

slope of secant line m=(0-(-3))/(2-(-1))=(0+3)/(2+1)=3/3=1

Next, solve for the y-intercept:

y=mx+b ........substitute coordinates (-1,-3) and m=1
-3=1*(-1)+b
-3=-1+b
-3+1=b
b=-2

the equation of the secant line is:
y=1*x-2
y=x-2




problem for you:
find the equation of the secant line through the points where x has the given values: f(x) = x^2 + 2x; x=3, x=5

 

Good. I will solve the "problem for you" later on today.

 

I must evaluate the given function at x = 3 and x = 5.
After doing so, I found two secant line points and they are (3, 15) & (5, 35).

I gotta find the slope of the secant line.

Let m_sec = slope of secant line.

m_sec = (35 - 15)/(5 - 3)

m_sec = 10

I now need the y-intercept, which is our b in the formula y = mx + b. I can use any of the two secant points to find b. I will use (3.15).

y = mx + b

15 = 10(3) + b

-15 = b.

The equation of the secant line is y = 10x - 15.

You say?

 

perfect

 

Thanks. This is the beginning of calculus. I fear Calculus lll. I've heard some nightmare stories from students about multivariable calculus.

 

I say that there is NOTHING in the original statement of the problem that requires you to find the equation of the line! When x= -1, \(y= (-1)^2- 4= -3\) and when x= 2, \(y= 2^2- 4= 0\) so this problem is just asking for the distance from (-1, -3) to (2, 0). That is \(\sqrt{(-1- 2)^2+ (-3- 0)^2}= \sqrt{(-3)^2+ (-3)^2}= 3\sqrt{2}\).

 

the original statement of the problem: Find the equation of the secant line that intersects the curve y = x^2 - 4 at x = -1 & x = 2.

 

Thank you.

 

A "secant line" to a curve is a line through two points on that curve. Here the curve is y= x^2- 4 and the two points are (-1, 1- 4)= (-1, -3) and (2, 4- 4)= (2, 0). Any (non-vertical) line can be written y= ax+ b.
-3= a(-1)+ b= b- a. 0= 2a+ b.

b= -2a so b- a= -2a- a= -3a= -3 so a= 1. Then b- a= b- 1= -3 and b= -3+ 1= -2.
The equation of the line is y= x- 2.

Check: If x= -1, y= -1- 2= -3. Yes, (-1, -3) is on that line. If x= 2, y= 2- 2= 0. Yes, (2, 0) is on that line.

(Checking is important! The first time I did this, I made a silly mistake that I only caught in checking.)

 

This post dates back to October 2021. Why reply now?