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3 years ago

Find the X intercepts of the equation f(x) = x^2 - 6x + 6

Mathematics

1 answer:

Arturiano [62]3 years ago

8 0

Answer:

x-intercepts = (x ≈ -0.87), (x ≈ 6.87)

Step-by-step explanation:

let x = 0, in equation for x-intercepts

solve for x using the

quadratic formula

x = −6 ± √ 36 + 24

= −6 ± √60

−2x

= −6 ± 2 √ 15

−2 = 3 ± √15

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What are the possible rational zeros of f(x) = x4 + 6x3 − 3x2 + 17x − 15?

aleksandr82 [10.1K]

Consider the polynomial $f(x) = x^4 + 6x^3 - 3x^2 + 17x- 15.$

The rational zeros could be only of form c/d, where c is integer number among divisors of the last term (-15) and d is integer number among divisors of the first term (1).

The divisors of -15 are: $\pm 1, \pm 3, \pm 5, \pm 15.$

The divisors of 1 are: $\pm 1.$

Possible rational zeros: $\pm 1, \pm 3, \pm 5, \pm 15.$

Check them:

$f(1)=1^4 + 6\cdot 1^3 - 3\cdot 1^2 + 17\cdot 1- 15=1+6-3+17-15=6\neq 0;$

$f(-1)=(-1)^4 + 6\cdot (-1)^3 - 3\cdot (-1)^2 + 17\cdot (-1)- 15=1-6-3-17-15=-40\neq 0;$

$f(3)=3^4 + 6\cdot 3^3 - 3\cdot 3^2 + 17\cdot 3- 15=81+162-27+51-15=252\neq 0;$

$f(-3)=(-3)^4 + 6\cdot (-3)^3 - 3\cdot (-3)^2 + 17\cdot (-3)- 15=81-162-27-51-15=-174\neq 0;$

$f(5)=5^4 + 6\cdot 5^3 - 3\cdot 5^2 + 17\cdot 5- 15=625+750-75+85-15=1370\neq 0;$

$f(-5)=(-5)^4 + 6\cdot (-5)^3 - 3\cdot (-5)^2 + 17\cdot (-5)- 15=625-750-75-85-15=-300\neq 0;$

$f(15)=15^4 + 6\cdot 15^3 - 3\cdot 15^2 + 17\cdot 15- 15=50625+20250-675+255-15=70440\neq 0;$

$f(-15)=(-15)^4 + 6\cdot (-15)^3 - 3\cdot (-15)^2 + 17\cdot (-15)- 15=50625-20250-675-255-15=29430\neq 0.$

There are no rational zeros.

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