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1 year ago

Find all x-intercepts and y-intercepts of the graph of the function. F(x)=x³-4x² - 4x + 16

Mathematics

1 answer:

Lady bird [3.3K]1 year ago

8 0

Answer: (0;16) (4;0) (2;0) (-2;0).

Step-by-step explanation:

Y-intercepts of the graph of the function F(x)=x³-4x² - 4x + 16:

$x=0\\F(x)=0^3-4*0^2-4*0+16\\F(x)=16.\\Hence,\ (0;16).$

X-intercepts of the graph of the function F(x)=x³-4x² - 4x + 16:

$F(x)=0\\0=x^3-4x^2-4x+16\\0=(x^3-4x^2)-(4x+16)\\0=x^2*(x-4)-4*(x-4)\\0=(x-4)*(x^2-4)\\0=(x-4)*(x^2-2^2)\\0=(x-4)*(x-2)*(x+2)\\x-4=0\\x_1=4.\\Hence,\ (4;0)\\x-2=0\\x_2=2.\\Hence,\ (2;0)\\x+2=0\\x_3=-2.\\Hence,\ (-2;0).$

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3 years ago

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2 years ago

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alex41 [277]

Given:

Radius of circle is 13 cm

Height of triangle is 5 cm

To find:

Value of 'a' and 'b'

Steps:

To find value of 'a', we will use Pythagoras theorem as the triangle is a right angle triangle,

A² + B² = C²

$a^{2} + 5^{2} = 13^{2}$

$a^{2} + 25 = 169$

$a^{2} = 169 - 25$

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$\sqrt{a^{2}}=\sqrt{144}$

$a = 12$

Now to find the value of 'b', i will use law of cosine,

$c=\sqrt{a^{2}+b^{2}-2ab(cos\beta ) }$

$12 = \sqrt{13^{2}+5^{2}-2(5)(13)(cos\beta )}\\12=\sqrt{169 + 25-130(cos\beta )}\\12=\sqrt{194-130(cos\beta )}\\144 = 194 - 130(cos\beta )\\50 = 130(cos\beta )\\cos\beta = 0.3846\\\beta = cos^{-1}(0.3846)\\\beta = 67.38$

Therefore, the values of 'a' and 'b' is 12 and 67.38 respectively

Happy to help :)

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