What is y=4x^2-9x+1 in vertex form

asked Jul 3, 2019 42.2k views

5 votes

by Telma

8.7k points

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7 votes

Answer:

$y = 4 \left(x - (9)/(8) \right)^2 - (65)/(16)$

Explanation:

y = 4x^2 - 9x + 1

You need to complete the square on the right side.

y = (4x^2 - 9x) + 1

y = 4(x^2 - (9)/(4)x) + 1

$y = 4 \left( x^2 - (9)/(4)x + \left( (9)/(8) \right)^2 \right) + 1 - 4\left( (9)/(8) \right)^2$

$y = 4 \left(x - (9)/(8) \right)^2 + 1 - 4 \left( (81)/(64) \right)$

$y = 4 \left(x - (9)/(8) \right)^2 + 1 - (81)/(16)$

$y = 4 \left(x - (9)/(8) \right)^2 + (16)/(16) - (81)/(16)$

$y = 4 \left(x - (9)/(8) \right)^2 - (65)/(16)$

Montells answered Jul 7, 2019

8.6k points

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