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What is the vertex form of #y= 8x^2+17x+1 #?

Answer 1

Avery Alexander

# y =8 (x + 17/16 )^2 - 257/32 #

The vertex form of the trinomial is; #y = a(x - h )^2 + k#

where the vertex's coordinates are (h, k).

the x-coordinate of the vertex is x # = -b/(2a)#

[from # 8x^2 + 17x + 1 #

b = 17, c = 1, and a = 8

so x-coord# = -17/16 #

and y-coord # = 8 xx (-17/16)^2 + 17 xx (-17/16) + 1 #

# = cancel(8) xx 289/cancel(256) - 289/16 + 1#

# = 289/32 - 578/32 + 32/32 = -257/32#

Find a point by using the formula: if x = 0, then y = 1, or (0,1)

and so : 1= a#(17/16)^2 -257/32 = (289a)/256 -257/32#

hence # a = (256 + 2056)/289 = 8#

equation is : # y = 8( x + 17/16)^2 - 257/32#

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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.