How do you write $y + 10 = 3 {(x - 1)}^{2}$ $y + 10 = 3(x - 1)^2$ in standard form?

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Answer 1 · 2018-03-28T01:13:23

$y = 3 x^{2} - 6 x - 7$ $y=3x^2-6x-7$

Given:

$y + 10 = 3 {(x - 1)}^{2}$ $y+10=3(x-1)^2$

Subtract $10$ $10$ from both sides.

$y = 3 {(x - 1)}^{2} - 10$ $y=3(x-1)^2-10$

Rewrite ${(x - 1)}^{2}$ $(x-1)^2$ as $(x - 1) (x - 1)$ $(x-1)(x-1)$ .

$y = 3 (x - 1) (x - 1) - 10$ $y=3(x-1)(x-1)-10$

$y = 3 (x^{2} - 2 x + 1) - 10$ $y=3(x^2-2x+1)-10$

Distribute the $3$ $3$ to the trinomial.

$y = 3 x^{2} - 6 x + 3 - 10$ $y=3x^2-6x+3-10$

Simplify.

$y = 3 x^{2} - 6 x - 7$ $y=3x^2-6x-7$

How do you write y + 10 = 3(x - 1)^2y+10=3(x−1)2y + 10 = 3(x - 1)^2 in standard form?