How do you express $y = 5 x^{2} - 3 x + 2$ $y=5x^2-3x+2$ in vertex form?

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How do you express $y = 5 x^{2} - 3 x + 2$ $y=5x^2-3x+2$ in vertex form?

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Answer 1 · 2016-07-04T19:39:25

$y = 5 {(x - \frac{3}{10})}^{2} + \frac{31}{20}$ $y = 5(x-3/10)^2+31/20$

Explanation:

$y = 5 x^{2} - 3 x + 2$ $y=5x^2-3x+2$

$= 5 (x^{2} - \frac{3}{5} x + \frac{2}{5})$ $=5(x^2-3/5x+2/5)$

$= 5 (x^{2} - \frac{3}{5} x + \frac{9}{100} - \frac{9}{100} + \frac{2}{5})$ $=5(x^2-3/5x+9/100-9/100+2/5)$

$= 5 {(x - \frac{3}{10})}^{2} + 5 (\frac{2}{5} - \frac{9}{100})$ $=5(x-3/10)^2+5(2/5-9/100)$

$= 5 {(x - \frac{3}{10})}^{2} + \frac{40 - 9}{20}$ $=5(x-3/10)^2+(40-9)/20$

$= 5 {(x - \frac{3}{10})}^{2} + \frac{31}{20}$ $=5(x-3/10)^2+31/20$

Bringing the two ends together, we have:

$y = 5 {(x - \frac{3}{10})}^{2} + \frac{31}{20}$ $y = 5(x-3/10)^2+31/20$

which is in vertex form:

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

with vertex $(h, k) = (\frac{3}{10}, \frac{31}{20})$ $(h, k) = (3/10, 31/20)$ and multiplier $a = 5$ $a=5$

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How do you express $y = 5 x^{2} - 3 x + 2$ $y=5x^2-3x+2$ in vertex form?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

How do you express y=5x2−3x+2y=5x^2-3x+2 in vertex form?

1 Answer

Explanation:

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How do you express $y = 5 x^{2} - 3 x + 2$ $y=5x^2-3x+2$ in vertex form?