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Graph f(x) =110(x+3)(x 1)(x 4) please explain graph f(x)=(x 3)(

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graph f(x) =1/10(x+3)(x-1)(x-4)
please explain
graph f(x)=(x-3)(x^2-x+1)
please explain
Solution
Let \'y\' = f(x) Hence y = (x+3)(x-1)(x-4) / 10
To make y = 0 then on or all the backeted terms must be
zero. So x+ 3 = 0 ; x = - 3 x - 1 = 0 ; x = 1 x - 4 = 0 ; x = 4
So -3,1,4 are the three points were the curve intercepts the
\'x\' axis. To find were the curve intercepts the \'y\' axis.
Make x = 0 Hence y = 3 * -1 * -4 / 10 y = 12/10 = 6/5 = 1
1/5 = 1.2 So the four interception points are (-3,0), (1,0),
(4,0) & (0, 6/5) When the three bracketed terms are
mulltiplied together the first terms is \'x^3\' Since \'x^3\' is
positive then from the left the curve rises, then falls, and
rises again. To find the max/min points use differential
calculus. y = (x^2 + 2x -3)(x - 4) /10 y = (x^3 + 2x^2 - 3x -
4x^2 - 8x + 12) /10 Collect like terms y = (x^3 - 2x^2 -11x +
12) / 10 Differentiate and equate to zero dy/dx = (3x^2 - 4x
- 11) / 10 = 0 x = {--4 +/- sqrt[(-4)^2 - 4(3)(-11)]} / 2(3) x =
{4 +/- sqrt[16 + 132]} / 6 x = { 4+/- sqrt[148]} / 6 x = {4 +/-
12.165525060596439377999368490404}/ 6 x=

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graph f(x) =1/10(x+3)(x-1)(x-4) please explain graph f(x)=(x-3)(x^2-x+1) please explain Solution Let \'y\' = f(x) Hence y = (x+3)(x-1)(x-4) / 10 To make y = 0 then on or all the backeted terms must be zero. So x+ 3 = 0 ; x = - 3 x - 1 = 0 ; x = 1 x - 4 = 0 ; x = 4 So -3,1,4 are the three points were the curve intercepts the \'x\' axis. To find were the curve intercepts the \'y\' axis. Make x = 0 Hence y = 3 * -1 * -4 / 10 y = 12/10 = 6/5 = 1 1/5 = 1.2 So the four interception points are ( -3,0), (1,0), (4,0) & (0, 6/5) When the three bracketed terms are mulltiplied together the first terms i ...
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