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What is the vertex form of #5y=-x^2 + 9x +8#?

Answer 1

Genesis Allen

#y=-1/5(x-9/2)^2+113/20#

We need the form of: #y="something"# so divide all of both sides by 5 giving:

#y=-1/5x^2+9/5x+8/5" ".......Equation(1)#

Write as:

#color(green)(y=-1/5(x^2-color(red)(9)x)+8/5)#

Halve the #color(red)(9)# and write as:

#color(green)(y=-1/5(x-color(red)(9)/2)^2+k+8/5)" "....Equation(2)#

The #k# is a correction factor as by doing the above you have added a value that is not in the original equation.

Set #color(green)(-1/5(-color(red)(9)/2)^2 +k=0)#

#=> k=+81/20#

Substitute for #k# in #Equation(2)# giving:

#color(green)(y=-1/5(x-color(red)(9)/2)^2+81/20+8/5)" "....Equation(2_a)#

#y=-1/5(x-9/2)^2+113/20#

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Answer 2

Nathan Axel

The vertex form of the quadratic equation 5y = -x^2 + 9x + 8 is y = -1/5(x - 9/2)^2 + 73/20.

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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.