How to find the y-intercept given $y = 4 {(x - 3)}^{2} - 1$ $y=4(x-3)^2-1$ ?

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Answer 1 · 2015-05-01T19:29:19

The y-intercept occurs at the point when $x = 0$ $x=0$

Given
$y = 4 {(x - 3)}^{2} - 1$ $y=4(x-3)^2 -1$

The y-intercept is at
$4 {(0 - 3)}^{2} - 1$ $4(0-3)^2-1$

$= 4 (9) - 1$ $=4(9) -1$

$= 35$ $=35$

How to find the y-intercept given y=4(x-3)^2-1y=4(x−3)2−1y=4(x-3)^2-1?