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Question

# Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.$y=-3x^2-9x-6$

Solution

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We will first write the equation in standard form. The standard form of a parabola is $y= a(x−h)^2+k$ where $(h,k)$ is the vertex.

$y=-3x^2-9x-6$

We will first transform the equation so that we can get the square of the binomial and so that we can later bring it to the standard form.

\begin{align*} y&= -3x^2-9x-\frac{27}{4}+\frac{3}{4} \\ y&= -3(x^2+3x+\frac{9}{4}) + \frac{3}{4}\\ y &= -3\times(x+\frac{3}{2})^2+\frac{3}{4} \end{align*}

We have obtained the standard form of the equation of the parabola: $y = -3\times(x+\frac{3}{2})^2+\frac{3}{4}$

$h$ $\longleftarrow$ $-\frac{3}{2}$ $k$ $\longleftarrow$ $\frac{3}{4}$

We will now easily determine the vertex:

$\text{Vertex}:(h,k)$

$\text{Vertex}:(-\frac{3}{2},\frac{3}{4})$

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