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uysha [10]
3 years ago
9

Which equation is y = 9x2 + 9x – 1 rewritten in vertex form?

Mathematics
1 answer:
Anastasy [175]3 years ago
7 0

Answer:

y = 9(x +\frac{1}{2}) ^ 2 -\frac{13}{4}

Step-by-step explanation:

An equation in the vertex form is written as

y = a (x-h) + k

Where the point (h, k) is the vertex of the equation.

 

For an equation in the form ax ^ 2 + bx + c the x coordinate of the vertex is defined as

x = -\frac{b}{2a}

In this case we have the equation y = 9x^2 + 9x - 1.

Where

a = 9\\\\b = 9\\\\c = -1

Then the x coordinate of the vertex is:

x = -\frac{9}{2(9)}\\\\x = -\frac{9}{18}\\\\x = -\frac{1}{2}

The y coordinate of the vertex is replacing the value of x = -\frac{1}{2} in the function

y = 9 (-0.5) ^ 2 + 9 (-0.5) -1\\\\y = -\frac{13}{4}

Then the vertex is:

(-\frac{1}{2}, -\frac{13}{4})

Therefore The encuacion excrita in the form of vertice is:

y = a(x +\frac{1}{2}) ^ 2 -\frac{13}{4}

To find the coefficient a we substitute a point that belongs to the function y = 9x^2 + 9x - 1

The point (0, -1) belongs to the function. Thus.

-1 = a(0 + \frac{1}{2}) ^ 2 -\frac{13}{4}

-1 = a(\frac{1}{4}) -\frac{13}{4}\\\\a = \frac{-1 +\frac{13}{4}}{\frac{1}{4}}\\\\a = 9

<em>Then the written function in the form of vertice is</em>

y = 9(x +\frac{1}{2}) ^ 2 -\frac{13}{4}

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