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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=3x29x6y=-3x^2-9x-6

Solution

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

y=3x29x6y=-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.

y=3x29x274+34y=3(x2+3x+94)+34y=3×(x+32)2+34\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×(x+32)2+34y = -3\times(x+\frac{3}{2})^2+\frac{3}{4}

hh \longleftarrow 32-\frac{3}{2} kk \longleftarrow 34\frac{3}{4}

We will now easily determine the vertex:

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

Vertex:(32,34)\text{Vertex}:(-\frac{3}{2},\frac{3}{4})

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