# What is the standard form of y= 4(x-5)^3 + 1 ?

Mar 6, 2017

$4 {x}^{3} - y - 61 = 0$

#### Explanation:

Standard form is $A x + B y + C = 0$

Using PEMDAS,

$y = 4 \left({x}^{3} - 15\right) + 1$

$y = 4 {x}^{3} - 60 + 1$

$y = 4 {x}^{3} - 61$

$y - y = 4 {x}^{3} - 61 - y$

$4 {x}^{3} - y - 61 = 0$