How do you write $y = 4 {(x - 3)}^{2} + 1$ $y = 4(x-3)^2+1$ in standard form?

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How do you write $y = 4 {(x - 3)}^{2} + 1$ $y = 4(x-3)^2+1$ in standard form?

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Answer 1 · 2018-07-02T05:44:53

$y = 4 x^{2} - 24 x + 37$ $y=4x^2-24x+37$

Explanation:

Given -

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

$y = 4 (x^{2} - 6 x + 9) + 1$ $y=4(x^2-6x+9)+1$

$y = 4 x^{2} - 24 x + 36 + 1$ $y=4x^2-24x+36+1$

$y = 4 x^{2} - 24 x + 37$ $y=4x^2-24x+37$

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How do you write $y = 4 {(x - 3)}^{2} + 1$ $y = 4(x-3)^2+1$ in standard form?

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How do you write $y = 4 {(x - 3)}^{2} + 1$ $y = 4(x-3)^2+1$ in standard form?