Question

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

Answer 1

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.

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

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

Then the written function in the form of vertice is

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