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3 years ago

15

The graph of f(x) = 2x3 – 19x2 + 57x – 54 is shown below. mc018-1.jpg How many roots of f(x) are rational numbers?

2 answers:

drek231 [11]3 years ago

8 0

There are three rational roots

BaLLatris [955]3 years ago

8 0

Answer: there are 3 root of f(x) which are rational numbers (2,0), (3,0) and (4.5,0)

Explanation:

Rational numbers are those that can be written in the form of p/q: q should not be equal to zero

since, the graph cuts x-axis at three points but rational are two

all integers are rational numbers

attached is the graph please have a look at it

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Which of the following is equivalent to the expression (9x^2y^3)(12x^-3y^5)(2xy)?

VladimirAG [237]

Answer:

$\large\boxed{(9x^2y^3)(12x^{-3}y^5)(2xy)=216y^9}\to\boxed{B.}$

Step-by-step explanation:

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3 years ago

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Find the set of values of k for which the line y=kx-4 intersects the curve y=x²-2x at 2 distinct points?

Sonbull [250]

Answer:

$-6 < k < 2$

Step-by-step explanation:

Given

$y = x^2 - 2x$

$y =kx -4$

Required

Possible values of k

The general quadratic equation is:

$ax^2 + bx + c = 0$

Subtract $y = x^2 - 2x$ and $y =kx -4$

$y - y = x^2 - 2x - kx +4$

$0 = x^2 - 2x - kx +4$

Factorize:

$0 = x^2 +x(-2 - k) +4$

Rewrite as:

$x^2 +x(-2 - k) +4=0$

Compare the above equation to: $ax^2 + bx + c = 0$

$a = 1$

$b= -2-k$

$c =4$

For the equation to have two distinct solution, the following must be true:

$b^2 - 4ac > 0$

So, we have:

$(-2-k)^2 -4*1*4>0$

$(-2-k)^2 -16>0$

Expand

$4 +4k+k^2-16>0$

Rewrite as:

$k^2 + 4k - 16 + 4 >0$

$k^2 + 4k - 12 >0$

Expand

$k^2 + 6k-2k - 12 >0$

Factorize

$k(k + 6)-2(k + 6) >0$

Factor out k + 6

$(k -2)(k + 6) >0$

Split:

$k -2 > 0$ or $k + 6> 0$

So:

$k > 2$ or k $> -6$

To make the above inequality true, we set:

$k < 2$ or $k >-6$

So, the set of values of k is:

$-6 < k < 2$

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Answer:

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Step-by-step explanation:

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