How do you write #f(x)= -2x^2+20x-49# in vertex form?

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Answer 1 · 2017-02-28T00:23:14

#y=-2(x-5)^2+1#

Standard Form equation: #y=ax^2+bx+c#
Vertex Form equation: #y=a(x-h)^2+k#

Where #a# is equal to the #a# value of the standard form equation and #(h,k)# is equal to the vertex of the equation.

In order to convert it, let's first fill in what we know. The #a# value of the given equation is -2.

So, we have:

#y=# -2 #(x-h^2)#+#k#

In order to find the vertex, you must use the equation #-b/(2a)#

Looking at the standard form equation, #b=20# and #a=-2#
So plugging in, you get #-20/(2*-2)#

Once solved, you're x value of your vertex is 5. Now, you plug 5 into your original standard form equation.

#y=-2(5)^2+20(5)-49#

Finally, your vertex is:

#(5, 1)#

Finally, plug it into your equation.
#y=-2(x-5)^2+1#