What is the vertex form of $y = 13 x^{2} + 3 x - 36$ $y=13x^2 +3x- 36$ ?

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What is the vertex form of $y = 13 x^{2} + 3 x - 36$ $y=13x^2 +3x- 36$ ?

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Answer 1 · 2016-02-01T04:27:20

vertex form: $y = {(x + \frac{3}{26})}^{2} - \frac{1881}{52}$ $y=(x+3/26)^2-1881/52$

Explanation:

1. Factor 13 from the first two terms.

$y = 13 x^{2} + 3 x - 36$ $y=13x^2+3x-36$

$y = 13 (x^{2} + \frac{3}{13} x) - 36$ $y=13(x^2+3/13x)-36$

2. Turn the bracketed terms into a perfect square trinomial.
When a perfect square trinomial is in the form $a x^{2} + b x + c$ $ax^2+bx+c$ , the $c$ $c$ value is ${(\frac{b}{2})}^{2}$ $(b/2)^2$ . Thus you divide $\frac{3}{13}$ $3/13$ by $2$ $2$ and square the value.

$y = 13 (x^{2} + \frac{3}{13} x + {(\frac{3}{13} x \div 2)}^{2}) - 36$ $y=13(x^2+3/13x+(3/13x-:2)^2)-36$

$y = 13 (x^{2} + \frac{3}{13} x + \frac{9}{676}) - 36$ $y=13(x^2+3/13x+9/676)-36$

3. Subtract 9/676 from the perfect square trinomial.
You cannot just add $\frac{9}{676}$ $9/676$ to the equation, so you must subtract it from the $\frac{9}{676}$ $9/676$ you just added.

$y = 13 (x^{2} + \frac{3}{13} x + \frac{9}{676}$ $y=13(x^2+3/13x+9/676$ $- \frac{9}{676}) - 36$ $color(red)(-9/676))-36$

4. Multiply -9/676 by 13.
The next step is to bring $- \frac{9}{676}$ $-9/676$ out of the brackets. To do this, multiply $- \frac{9}{676}$ $-9/676$ by the $a$ $a$ value, $13$ $13$ .

$y = 13 (x^{2} + \frac{3}{13} x + \frac{9}{676}) - 36 [(- \frac{9}{676}) \cdot (13)]$ $y=color(blue)13(x^2+3/13x+9/676)-36[color(red)((-9/676))*color(blue)((13))]$

5. Simplify.

$y = (x^{2} + \frac{3}{13} x + \frac{9}{676}) - 36 - \frac{9}{52}$ $y=(x^2+3/13x+9/676)-36-9/52$

$y = (x^{2} + \frac{3}{13} x + \frac{9}{676}) - \frac{1881}{52}$ $y=(x^2+3/13x+9/676)-1881/52$

6. Factor the perfect square trinomial.
The last step is to factor the perfect square trinomial. This will allow you to determine the coordinates of the vertex.

$y = {(x + \frac{3}{26})}^{2} - \frac{1881}{52}$ $color(green)(y=(x+3/26)^2-1881/52)$

$:.$ , the vertex form is $y=(x+3/26)^2-1881/52$ .

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What is the vertex form of $y = 13 x^{2} + 3 x - 36$ $y=13x^2 +3x- 36$ ?

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What is the vertex form of y=13x^2 +3x- 36 y=13x2+3x−36y=13x^2 +3x- 36 ?

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What is the vertex form of $y = 13 x^{2} + 3 x - 36$ $y=13x^2 +3x- 36$ ?