How do you rewrite $y = 3 x^{2} - 24 x + 10$ $y = 3x^2 - 24x + 10$ in vertex form?

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Answer 1 · 2017-07-15T13:43:03

The vertex form is $y = 3 {(x - 4)}^{2} - 38$ $y=3(x-4)^2-38$

The vertex form of a parabola is

$y = a {(x - h)}^{2} + k$ $y=a(x-h)^2+k$

Here, we have

$y = 3 x^{2} - 24 x + 10$ $y=3x^2-24x+10$

$y = 3 (x^{2} - 8 x) + 10$ $y=3(x^2-8x)+10$

We complete the square by adding half of the coefficient of $x$ $x$ squared and substract this value.

$y = 3 (x^{2} - 8 x + 16 - 16) + 10$ $y=3(x^2-8x+16-16)+10$

$y = 3 (x^{2} - 8 x + 16) + 10 - 48$ $y=3(x^2-8x+16)+10-48$

$y = 3 {(x - 4)}^{2} - 38$ $y=3(x-4)^2-38$

The vertex is $= (h, k) = (4, - 38)$ $=(h,k)=(4,-38)$

graph{3(x-4)^2-38 [-42.64, 39.6, -46.42, -5.33]}

How do you rewrite y=3x2−24x+10y = 3x^2 - 24x + 10 in vertex form?