Question

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

Answer 1

Answer:

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

Step-by-step explanation:

$y = 4x^2 - 9x + 1$

You need to complete the square on the right side.

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

$y = 4(x^2 - \dfrac{9}{4}x) + 1$

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

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

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

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

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