Find Equation of the Secant Line

[nycmath]

Precalculus
Michael Sullivan
Edition 10
Chapter 2, Section 2.3

74. h(x) = -2x^2 + x

Find an equation of the secant line containing (0, h(0)) and
(3, h(3)).

[Guest]

h(x)= -2[tex]x^{2 } +x[/tex]

We are looking for an equation of a line that intersects the given function[tex]\rightarrow[/tex] h(x)

The desired equation passes through points A(0 ;0) and B(3 ; -15) (1) .

The simple equation of a line has the form : y= kx +n (2)

We search k=? n=?
Replace (1) in (2) .
[tex]\begin{array}{|l} 0 = k.0+n \\ - 15 = k.3+n \end{array}[/tex]

[tex]\begin{array}{|l} n = 0 \\ -15= 3k+0 \end{array}[/tex]

[tex]\begin{array}{|l} n = 0 \\ k = - 5 \end{array}[/tex]

Then we go back to (2) y= -5x+0

Answer y= -5x

[nycmath]

h(x)= -2[tex]x^{2 } +x[/tex]

We are looking for an equation of a line that intersects the given function[tex]\rightarrow[/tex] h(x)

The desired equation passes through points A(0 ;0) and B(3 ; -15) (1) .

The simple equation of a line has the form : y= kx +n (2)

We search k=? n=?
Replace (1) in (2) .
[tex]\begin{array}{|l} 0 = k.0+n \\ - 15 = k.3+n \end{array}[/tex]

[tex]\begin{array}{|l} n = 0 \\ -15= 3k+0 \end{array}[/tex]

[tex]\begin{array}{|l} n = 0 \\ k = - 5 \end{array}[/tex]

Then we go back to (2) y= -5x+0

Answer y= -5x

Very cool.

[Eigenvalue]

1. Find the slope: [tex]\frac{f(3)-f(0)}{3}[/tex]
=[tex]\frac{-2(3)²+3}{3}[/tex]
=[tex]\frac{-15}{3}[/tex]=-5

2. Use the points to find an equation:
y-0=-5(x-0)
y=-5x

3. Check: is (3, -15) on the line?
-15=-5(3)

[nycmath]

1. Find the slope: [tex]\frac{f(3)-f(0)}{3}[/tex]
=[tex]\frac{-2(3)²+3}{3}[/tex]
=[tex]\frac{-15}{3}[/tex]=-5

2. Use the points to find an equation:
y-0=-5(x-0)
y=-5x

3. Check: is (3, -15) on the line?
-15=-5(3)

Very good answer. I like the ending question.