Whats the answer for this????

Question

Let f(x)=9.4x2−4.6xf(x)=9.4x2-4.6x. Find the equation of the secant line on the graph of f(x)f(x) between x1=3x1=3 and x2=12x2=12. Write your answer in mx+bmx+b format.

Answer 1

f(x) = 9.4x^2- 4.6x

= 9.4(x^2 - 4.6x/9.4 + (2.3/9.4)^2) - 2.3^2/9.4

= 9.4(x-2.3/9.4)^2 - 5.29/9.4

is an upward opening parabola with vertex at (2.3/9.4, 5.29/9.4) = about ( 0.24,0.56)

x1 =3, x2 =12

it helps to sketch the parabola and plot the points (3,f(3)) and (12,f(12)

connect the two points with a straight line, that's the secant line

f(3) = 9.4(9)-4.6(3) = 84.6-13.8 = 70.8

f(12) = 9.4(144)-4.6(12) = 1363.6- 55.2 = 1308.4

slope of the secant line = (1308.4-70.8)/(12-3) = 1237.6/9 = 137.5111...

equation of the secant line is

slope = (y-70.8)/(x-3) = 1237.6/9

y = (1237.6/9)x - 1237.6/3 + 70.8

y= (1237.6/9)x - 1025.2/3

y = about 137.5x - 341.7

or

check the slope calculation

f(x) = 9.4x^2 -4.6x

f'(x) = 18.8x -4.6

average of 3 and 12 = 7.5

f'(7.5) = 18.8(7.5)-4.6 = 136.4 = close to the slope of the secant line

Answer 2

Use two-point form fro Agebra 1:

(y - f(3)) / (x - 3) = (f(12) - f(3)) / (12 - 3)