Find the product. a 4(3a2 - 2a + 1) (typed answers not selection)

Mathematics

1 answer:

Rus_ich [418]3 years ago

6 0

Answer:

$3a^6-2a^5+a^4$

Step-by-step explanation:

Use the rules of indices to find the product.

One of the rules is that when we multiply indices that have the same base we add the powers. For example:

$x^4 * x^5 = x^9$

So to work out $a^4(3a^2-2a+1)$ we need to follow this rule.

Let's expand the bracket:

$a^4(3a^2-2a+1) = (a^4)(3a^2)+(a^4)(-2a)+(a^4)(1)$

Let's simplify keeping in mind the rule we know.

$(a^4)(3a^2)=3a^6$

$(a^4)(-2a) = -2a^5$ - remember that $a = a^1$

$(a^4)(1) = a^4$

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(01.03. 106, 1.08 HC)

zlopas [31]

Answer:

A) see attached for a graph. Range: (-∞, 7]

B) asymptotes: x = 1, y = -2, y = -1

C) (x → -∞, y → -2), (x → ∞, y → -1)

Step-by-step explanation:

A graphing calculator is useful for graphing the function. We note that the part for x > 1 can be simplified:

$\dfrac{-x^2+x+2}{x^2-3x+2}=-\dfrac{(x-2)(x+1)}{(x-2)(x-1)}=-\dfrac{x+1}{x-1}\quad x\ne 2$

This has a vertical asymptote at x=1, and a hole at x=2.

The function for x ≤ 1 is an ordinary exponential function, shifted left 1 unit and down 2 units. Its maximum value of 3^-2 = 7 is found at x=1.

The graph is attached.

The range of the function is (-∞, 7].

As we mentioned in Part A, there is a vertical asymptote at x = 1. This is where the denominator (x-1) is zero.

The exponential function has a horizontal asymptote of y = -2; the rational function has a horizontal asymptote of y = (-x/x) = -1. The horizontal asymptote of the exponential would ordinarily be y=0, but this function has been translated down 2 units.

The end behavior is defined by the horizontal asymptotes:

for x → -∞, y → -2

for x → ∞, y → -1