What is the vertex form of $5 y = 11 x^{2} - 15 x - 9$ $5y = 11x^2-15x-9$ ?

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What is the vertex form of $5 y = 11 x^{2} - 15 x - 9$ $5y = 11x^2-15x-9$ ?

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Answer 1 · 2017-04-10T05:05:28

$y = \frac{11}{5} {(x - \frac{15}{22})}^{2} - \frac{621}{220}$ $y=11/5(x-15/22)^2-621/220$

Explanation:

Vertex form of such equation is $y = a {(x - h)}^{2} + k$ $y=a(x-h)^2+k$ , with $(h, k)$ $(h,k)$ as vertex.

Here we have $5 y = 11 x^{2} - 15 x - 9$ $5y=11x^2-15x-9$

or $y = \frac{11}{5} x^{2} - 3 x - \frac{9}{5}$ $y=11/5x^2-3x-9/5$

or $y = \frac{11}{5} (x^{2} - 3 \times \frac{5}{11} x) - \frac{9}{5}$ $y=11/5(x^2-3xx5/11x)-9/5$

$= \frac{11}{5} (x^{2} - 2 \times \frac{15}{22} x + {(\frac{15}{22})}^{2} - {(\frac{15}{22})}^{2}) - \frac{9}{5}$ $=11/5(x^2-2xx15/22 x+(15/22)^2-(15/22)^2)-9/5$

$= \frac{11}{5} {(x - \frac{15}{22})}^{2} - {(\frac{15}{22})}^{2} \times \frac{11}{5} - \frac{9}{5}$ $=11/5(x-15/22)^2-(15/22)^2xx11/5-9/5$

$= \frac{11}{5} {(x - \frac{15}{22})}^{2} - \frac{45}{44} - \frac{9}{5}$ $=11/5(x-15/22)^2-45/44-9/5$

$= \frac{11}{5} {(x - \frac{15}{22})}^{2} - \frac{45 \times 5 + 44 \times 9}{220}$ $=11/5(x-15/22)^2-(45xx5+44xx9)/220$

$= \frac{11}{5} {(x - \frac{15}{22})}^{2} - \frac{225 + 396}{220}$ $=11/5(x-15/22)^2-(225+396)/220$

$= \frac{11}{5} {(x - \frac{15}{22})}^{2} - \frac{621}{220}$ $=11/5(x-15/22)^2-621/220$

and vertex is $(\frac{15}{22}, - \frac{621}{220})$ $(15/22,-621/220)$

graph{5y=11x^2-15x-9 [-4.667, 5.333, -4.12, 0.88]}

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What is the vertex form of $5 y = 11 x^{2} - 15 x - 9$ $5y = 11x^2-15x-9$ ?

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