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Anna35 [415]
3 years ago
13

Write a cubic function whose graph goes through the points: (-5, 0) (1, 0) (2, -2) (4, 0)

Mathematics
1 answer:
Free_Kalibri [48]3 years ago
5 0

Answer:

f(x) = \frac{1}{7}(x+5)(x-1)(x-4)

Step-by-step explanation:

Let the equation of the give cubic function is,

f(x) = p(x - a)(x - b)(x - c)

Here, a, b, c and d are the x-intercepts of the given graph.

Since, x-intercepts given in the graph are x = -5, 1 and 4,

Equation of the curve will be,

f(x) = p(x + 5)(x - 1)(x - 4)

Since, the graph of this function passes through (2, -2) also

-2 = p(2 + 5)(2 - 1)(2 - 4)

-2 = -p(14)

p = \frac{2}{14}

p = \frac{1}{7}

Therefore, equation will be,

f(x) = \frac{1}{7}(x+5)(x-1)(x-4)

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wel

Given:

The figure of rectangle.

To find:

a. The diagonal of the rectangle.

b. The area of the rectangle.

c. perimeter of the rectangle.

Solution:

(a)

In a right angle triangle,

\sin \theta=\dfrac{Perpendicular}{Hypotenuse}

\sin 30=\dfrac{12}{Hypotenuse}

\dfrac{1}{2}=\dfrac{12}{Hypotenuse}

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So, the diagonal of the of the rectangle is 24 units.

(b)

In a right angle triangle,

\tan \theta=\dfrac{Perpendicular}{Base}

\tan 30=\dfrac{12}{Base}

\dfrac{1}{\sqrt{3}}=\dfrac{12}{Base}

Base=12\sqrt{3}

Length of the rectangle is 12 and width of the rectangle is 12\sqrt{3}. So, the area of the rectangle is:

Area=length \times width

Area=12 \times 12\sqrt{3}

Area=144\sqrt{3}

So, the area of the rectangle is 144\sqrt{3} sq. units.

(c)

Perimeter of the rectangle is:

P=2(length+width)

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