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What is the vertex form of #y= (5x+2)^2 + 11x(5x+2)+30#?

Answer 1

Isaiah Anthony

#y=80(x^2+21/80)^2+2279/80#

First, let's make this simpler.

#y=(5x+2)^2+11x(5x+2)+30#

#=25x^2+20x+4+55x^2+22x+30#

#=80x^2+42x+34#

#=80(x^2+42/80x)+34#

#=80(x^2+2xx21/80x+(21/80)^2-(21/80)^2)+34#

#=80(x^2+21/80)^2-(21/80)^2xx80+34#

#=80(x^2+21/80)^2-441/80+34#

#=80(x^2+21/80)^2+2279/80#

which is in vertex form and vertex is #(-21/80,2279/80)# or #(-21/80,28 39/80)# and graph appears as follows:

graph{[-2, 2, -10.9, 149.1]} = 80x^2 + 42x + 34

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Answer 2

Savannah Anderson

#y=80(x^2+21/80)^2+2279/80#

First, let's make this simpler.

#y=(5x+2)^2+11x(5x+2)+30#

#=25x^2+20x+4+55x^2+22x+30#

#=80x^2+42x+34#

#=80(x^2+42/80x)+34#

#=80(x^2+2xx21/80x+(21/80)^2-(21/80)^2)+34#

#=80(x^2+21/80)^2-(21/80)^2xx80+34#

#=80(x^2+21/80)^2-441/80+34#

#=80(x^2+21/80)^2+2279/80#

which is in vertex form and vertex is #(-21/80,2279/80)# or #(-21/80,28 39/80)# and graph appears as follows:

graph{[-2, 2, -10.9, 149.1]} = 80x^2 + 42x + 34

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Answer 3

Kennedy Adams

The vertex form of the given expression (y = (5x+2)^2 + 11x(5x+2) + 30) is (y = 5(x + \frac{11}{10})^2 + \frac{197}{4}).

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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.