What is the standard form of #y= (3x – 4) (x^2+16) #?

Answer 1

See a solution process below:

To put this equation in standard form we must multiply the two terms on the right side of the equation by multiplying each individual term in the left parenthesis by each individual term in the right parenthesis.

#y = (color(red)(3x) - color(red)(4))(color(blue)(x^2) + color(blue)(16))# becomes:

#y = (color(red)(3x) xx color(blue)(x^2)) + (color(red)(3x) xx color(blue)(16)) - (color(red)(4) xx color(blue)(x^2)) - (color(red)(4) xx color(blue)(16))#

#y 3x^3 + 48x - 4x^2 - 64#

Now, put the terms in order by largest exponent on the left to the lowest exponent or constant on the right:

#y 3x^3 - 4x^2 + 48x - 64##

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Answer 2

Abigail Allen

The standard form of ( y = (3x - 4)(x^2 + 16) ) is ( y = 3x^3 - 12x^2 + 48x - 64 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.