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2 years ago

URGENT PLEASE DUE TONIGHT ;(( functions

Mathematics

1 answer:

Montano1993 [528]2 years ago

5 0

The value of (f/g)(-1) is 0, value of (g . f) (2) is -3√3, the value of (g - f)(-1) is √15, and the value of (g + f) (2) is √3 - 3

<h3>What is a function?</h3>

It is defined as a special type of relationship, and they have a predefined domain and range according to the function every value in the domain is related to exactly one value in the range.

We have two functions:

f(x) = 1 - x²

g(x) = √(11 - 4x)

f(-1) = 0

g(-1) = √15

f(2) = -3

g(2) = √3

(f/g)(-1) = f(-1)/g(-1) = 0/√15 = 0

(g . f) (2) = g(2)f(2) = (√3)(-3) = -3√3

(g - f)(-1) = g(-1) - f(-1) = √15 - 0 = √15

(g + f) (2) = g(2) + f(2) = √3 + (-3) = √3 - 3

Thus, the value of (f/g)(-1) is 0, value of (g . f) (2) is -3√3, the value of (g - f)(-1) is √15, and the value of (g + f) (2) is √3 - 3

Learn more about the function here:

brainly.com/question/5245372

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Answer:

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Answer:

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Step-by-step explanation:

Given:

$y = -6x^2 + 3x + 2$

Required

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The vertex form of an equation is in form of: $y = a(x - h)^2+ k$

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Get a perfect square of coefficient of x; then add to both sides

------------------------------------------------------------------------------------

<em>Rough work</em>

The coefficient of x is $\frac{-1}{2}$

It's square is $(\frac{-1}{2})^2 = \frac{1}{4}$

Adding inside the bracket of $-6(x^2 - \frac{x}{2})$ to give: $-6(x^2 - \frac{x}{2} + \frac{1}{4})$

To balance the equation, the same expression must be added to the other side of the equation;

Equivalent expression is: $-6(\frac{1}{4})$

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The expression becomes

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$y - 2 -\frac{6}{4}= -6(x^2 - \frac{x}{2} + \frac{1}{4})$

$y - 2 -\frac{3}{2}= -6(x^2 - \frac{x}{2} + \frac{1}{4})$

Factorize the expression on the right hand side

$y - 2 -\frac{3}{2}= -6(x - \frac{1}{2})^2$

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$y +\frac{7}{2} = -6(x - \frac{1}{2})^2$

Make y the subject of formula

$y = -6(x - \frac{1}{2})^2 -\frac{7}{2}$

<em>Solved</em>

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