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write the function in vertex form.
y=4x2 - 8x + 5

Sagot :

eternal3fear
you will have to complete the square
factor the 4 out of the first two terms
y = 4(x^2 - 2x) + 5

x^2 - 2x + 1 = (x - 1)^2
we want what we have to look like this
the difference between (x^2 - 2x) and (x^2 - 2x + 1) is the +1
so we add a 1 and subtract a 1 to keep balance

y = 4(x^2 - 2x + 1  -1) + 5
y = 4(x^2 - 2x + 1) - 4 + 5   [moved the extra -1 out of it and distributive property]
y = 4(x - 1)^2 + 1   [factored and added like terms]

y = 4(x-1)^2 + 1
or
y - 1 = 4(x-1)^2

is now in vertex form.
favour14

a=4, b=-8, c=5

h=-b/2a= -(-8)/2(4)

8/8=h=1

K= a(h)^2+b(h)+c

K= 4(1)^2-8(1)+5

K= 4-8+5

K= 1, Vertex  Form: a(x-h)^2+k

Vertex Form: Y= 4(X-(1)^2+K

Y= 4(x-1)^2+1

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