Let f(x) = 4x^3 - 3x - 1?

Question

Please Help Me
1) find the x-intercepts of the graph of f
2) write an equation for the tangent line to the graph of f at x = 2
3) Write an equation of the graph that is the reflection across the y-axis of the graph of f

Answer 1

1.
Set y = 0 to find the x-intercepts
0 = 4x^3 - 3x - 1
0 = (4x^3 - 4x) + (x - 1)
0 = 4x(x^2 - 1) + (x - 1)
0 = 4x(x - 1)(x + 1) + (x - 1)
0 = (x - 1)(4x(x + 1) + 1)
0 = (x - 1)(4x^2 + 4x + 1)
0 = (x - 1)(2x + 1)(2x + 1)
x = 1, -1/2
x-intercepts: (1, 0), (-1/2, 0)

2.
Take the derivative and plug in x=2 to find the slope of the tangent line at x=2
f'(x) = 12x^2 - 3 + 0
= 12x^2 - 3
f'(2) = 12(2^2) - 3
= 48 - 3
= 45

Now find the y-value at x=2
f(2) = 32 - 6 - 1
= 25

Use point-slope formula to find the equation
y - 25 = 45(x - 2)
y - 25 = 45x - 90
y = 45x - 65

3.
To reflect it across the y-axis, simply change the sign of all the coefficients of terms with odd exponents
f(x) = -4x^3 + 3x - 1

Answer 2

Hi,

1) The x-intercepts are at x = -½ and at x = 1. These can be gotten from a graphing calculator. You can also factor this into f(x) = (2x + 1)(2x + 1)(x - 1) and solve each factor to get these intercepts. You can also do synthetic division by the possible rational roots of ±¼, ±½, and ±1 to find the only remainders of zero occur when you divide by -½ and 1. (Remember on synthetic division to fill in a 0 for the missing x² term.)

2) The first derivative of f(x) is f'(x) = 12x² - 3. When x = 2, this equals 12(2)² - 3 = 48 - 3 = 45. 45 is the slope of the tangent when x = 2. The point on f(x) when x = 2 is (2,25). The equation of the tangent there in point-slope form is y - 25 = 45(x - 2). In slope intercept form this is y = 45x - 65

3) The equation of the graph that is the reflection across the y-axis of the graph of f(x) is g(x) = -4x³ + 3x - 1. This equation comes from replacing x in the original equation with (-x) and simplifying the equation.
f(x) = 4x³ - 3x - 1
g(x) = 4(-x)³ - 3(-x) - 1
g(x) = -4x³ + 3x - 1

I hope that helps!! :-)

Answer 3

hi, = = = = = = = = = = = = = = 4x³ – 3x – a million = 0 ??? word that 4–3–a million=0 4x³ – 4x + x – a million = 0 ??? split -3=-4+a million 4x(x² – a million) + (x – a million) = 0 ??? ingredient by skill of grouping 4x(x – a million)(x + a million) + (x – a million) = 0 ??? simply by fact a²–b²=(a–b)(a+b) (x – a million)(4x² + 4x + a million) = 0 ??? ingredient out by skill of x–a million (x – a million)(2x + a million)² = 0 ??? simply by fact a²+2ab+b²=(a+b)² for this reason the sole roots are: x=a million and x=-½ and that they are the two rational! = = = = = = = = = = = = = = 2x³ + 8x² + 7x + 3 = 0 2x³ + 6x² + 2x² + 6x + x + 3 = 0 2x²(x + 3) + 2x(x + 3) + (x + 3) = 0 ??? ingredient by skill of grouping (x + 3)(2x² + 2x + a million) = 0 (x + 3)(4x² + 4x + 2) = 0 ??? Multiply by skill of two (x + 3)(4x² + 4x + a million + a million) = 0 (x + 3)[(2x + a million)² + a million] = 0 (x + 3)[(2x + a million)² – i²] = 0 ??? simply by fact i²=-a million (x + 3)(2x + a million + i)(2x + a million – i) = 0 ??? simply by fact a²–b²=(a+b)(a–b) for this reason the three roots are: x = -3 and x = -(a million ± i)/2 = = = = = = = = = = = = = = in accordance to the rational root theorem, ±3 won't be able to be root of x³ + 3x² – 4x + 2 for this reason the fact is fake. Regards, Dragon.Jade :-)

Answer 4

Well obviously one x-intercept is (1,0) since 4-3-1=0.
Now you can divide 4x^3 -3x-1 by x-1 getting 4x^2+4x+1. This factors to (2x+1)^2, so x= -1/2 is a double root.
So x- intercept at (-1/2,0) and (1,0)

dy/dx = 12x^2-3.
When x = 2 dy/dx = 45 so tangent is y= 45x +b.
When x = 2, f(x) = 25, so 25=45*2+b --> b= -65.
So equation of tangent is y= 45x -65

y= -4x^3 +3x-1 is efletion across y-axis. Just replace x with -x

Answer 5

find the roots by pluging the factors of the constants or the factors of the leading coefficient.

try 1

f(1) = 4(1)^3 - 3(1) - 1
f(1) = 4 - 3 - 1
f(1) = 0

1 is one of the root, so (x - 1) is one of the factors.

sythetic long division. Since x - 1 is in the from x - k, k = 1

f(x) = 4x^3 + 0x^2 - 3x - 1

1 l 4 ... 0 ... -3 ... -1
...l ..... 4 ..... 4 ..... 1
-------------------------
.... 4 .. 4 .... 1 .... 0

f(x) = (x - 1) (4x^2 + 4x + 1)


factor the rest
f(x) = (x - 1) (2x + 1)^2

so the x-int are (1,0) and (-1/2,0)


2) (2, f(2)) ==> (2,25)

find the deriative


f'(x) = 12x^2 - 3

m = 12(2)^2 - 3
m = 45

y - y1 = m(x - x1)
y - 25 = 45(x - 2)
y = 45(x - 2) + 25
y = 45x - 90 + 25
y = 45x - 65


3) when the graph reflects the y-asix, the x changes
f(x) = -4x^3 + 3x - 1

Answer 6

4x^3 - 3x - 1=(x-1)(4x^2 +4x +1)=(x -1 )(2x+1)^2
1) f(x) = 0 then x=1 or x=-1/2 i.e. (1,0) and ( - 1/2 ,0)
2)f'(x) =12x^2-3 = 45 =slope and f(2)=32-6-1=25
then y -25/x - 2 = 45 then the eqn.is 45x - y - 65=0
3) as when f(x) =f(-x) then it is even function i.e. it graph is symmetric on the Y-axis then f( -x) = - 4x^3 + 3 x - 1