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3 years ago

Graphing Quadratic Functions (help please )

Mathematics

1 answer:

Reil [10]3 years ago

6 0

Answer:

Zeroes: x = 4, x = -6

The vertex is at ( -1 , 50 )

Step-by-step explanation:

I'll do the first one, you try the rest

All of these are written in y = a(x - x1)(x - x2) where x1 and x2 are given and x are the two different zeroes (or x-intercepts)

f(x) = -2(x-4)(x+6)

The zeroes are x - 4 = 0 and x + 6 = 0 if you solve, you get 4 and -6

Zeroes: x = 4, x = -6

The x value of the vertex can be found using -b/2a (in the form ax^2 + bx + c)

First expand f(x)

-2(x^2 + 2x - 24) => -2x^2 - 4x + 48

b = -4 and a = -2

-(-4)/2(-2) = 4/-4 = -1

the x-value of the vertex is -1

Now substitute -1 as x to find the y-value of the vertex

-2(-1-4)(-1+6) = -2(-5)(5) = 50 <= y-value of vertex

The vertex is at ( -1 , 50 )

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we know that

A polynomial in the form $a^{3} +b^{3}$ is called a sum of cubes

Let's verify each case to determine the solution

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we know that

$-64=-4^{3}$

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therefore

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case B) $-32x^{6} y^{12} +125x^{16} y^{3}$

we know that

$-32=-2^{5}$ -------> is not a perfect cube

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$y^{12}= (y^{4})^{3}$

$125=5^{3}$

$x^{16}=x^{15} *x=x*(x^{5})^{3}$ -------> is not a perfect cube

$y^{3}= (y)^{3}$

therefore

the case B) is not a sum of cubes

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we know that

$32=2^{5}$ -------> is not a perfect cube

$x^{6}= (x^{2})^{3}$

$y^{12}= (y^{4})^{3}$

$125=5^{3}$

$x^{9}=(x^{3})^{3}$

$y^{3}= (y)^{3}$

therefore

the case C) is not a sum of cubes

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we know that

$64=4^{3}$

$x^{6}= (x^{2})^{3}$

$y^{12}= (y^{4})^{3}$

$125=5^{3}$

$x^{9}=(x^{3})^{3}$

$y^{3}= (y)^{3}$

Substitute

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$(4x^{2}y^{4})^{3} +(5x^{3}y)^{3}$

therefore

the answer is

$64x^{6} y^{12} +125x^{9} y^{3}$ is a sum of cubes

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