What is the vertex form of #5y = 11x^2-15x-9#?

Answer 1

Julia Adams

#y=11/5(x-15/22)^2-621/220#

Vertex form of such equation is #y=a(x-h)^2+k#, with #(h,k)# as vertex.

Here we have #5y=11x^2-15x-9#

or #y=11/5x^2-3x-9/5#

or #y=11/5(x^2-3xx5/11x)-9/5#

#=11/5(x^2-2xx15/22 x+(15/22)^2-(15/22)^2)-9/5#

#=11/5(x-15/22)^2-(15/22)^2xx11/5-9/5#

#=11/5(x-15/22)^2-45/44-9/5#

#=11/5(x-15/22)^2-(45xx5+44xx9)/220#

#=11/5(x-15/22)^2-(225+396)/220#

#=11/5(x-15/22)^2-621/220#

and vertex is #(15/22,-621/220)#

graph{5y=11x^2-15x-9 [-4.667, 5.333, -4.12, 0.88]}

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Answer 2

Nathan Axel

The vertex form of the quadratic equation 5y = 11x^2 - 15x - 9 is:

( y = \frac{11}{5}(x - h)^2 + k ),

where ( h ) and ( k ) are the coordinates of the vertex.

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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.