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3 years ago

Where r the turning points of x^2(x-4)^7(x+3)^6

Mathematics

1 answer:

Tomtit [17]3 years ago

8 0

Answer:

There are 5 turning points at x = -3, -1.11, 0, 1.44 and 4.

Step-by-step explanation:

The product rule for 3 terms is

F' (x) = f'(x) g(x) h(x) + f(x) g'(x) h(x) + f(x) g(x) h'(x)

F(x) = x^2(x - 4)^7 (x + 3)^6

F' (x) = 2x(x-4)^7(x+3)^6 + x^2 7(x - 4)^6 (x + 3)^6 + x^2(x-4)^7 6(x+3)^5 = 0 at the turning points.

That is some equation!!

By observation, 3 roots are x = 0 , x = 4 and x = -3 so there are turning points at these values of x.

I'll use software to graph it to find any more real solutions - they are -1.11 and 1.44 to nearest hundredth.

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3 years ago

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densk [106]

Answer:

$A.\ y = x^2 - 6x + 13$ is the correct answer.

Step-by-step explanation:

We know that vertex equation of a parabola is given as:

$y = a(x-h)^2+k$

where $(h,k)$ is the vertex of the parabola and

$(x,y)$ are the coordinate of points on parabola.

As per the question statement:

The parabola opens upwards that means coefficient of $x^{2}$ is positive.

Let $a = +1$

Minimum of parabola is at x = 3.

The vertex is at the minimum point of a parabola that opens upwards.

$\therefore$ $h = 3$

Putting value of a and h in the equation:

$y = 1(x-3)^2+k\\\Rightarrow y = (x-3)^2+k\\\Rightarrow y = x^2-6x+9+k$

Formula used: $(a-b)^2=a^{2} +b^{2} -2\times a \times b$

Comparing the equation formulated above with the options given we can observe that the equation formulated above is most similar to option A.

Comparing $y = x^2 - 6x + 13$ and $y = x^2-6x+9+k$

13 = 9+k

k = 4

Please refer to the graph attached.

Hence, correct option is $A.\ y = x^2 - 6x + 13$

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Step-by-step explanation:

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