Exercise 9 Page 224

We have a quadratic function written in standard form, and we want to rewrite it in vertex form. Standard Form y= ax^2+ bx+c Given Equation y=- 3x^2-6x-8 ⇔ y= - 3x^2+( - 6)x+(- 8) In the given equation, a= - 3, b= - 6, and c=- 8. Let's now recall the vertex form of a quadratic function. Vertex Form: y=a(x-h)^2+k In this equation, a is the leading coefficient of the quadratic function, and the point (h,k) is the vertex of the parabola. Also, note that a vertical line with the equation x=h is the axis of symmetry. By substituting our given values for a and b into the expression - b2a, we can find h.

- b/2a

SubstituteII

a= - 3, b= - 6

- - 6/2( - 3)

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Simplify

MultPosNeg

a(- b)=- a * b

- - 6/- 6

DivNegNeg

- a/- b=a/b

-6/6

QuotOne

a/a=1

- 1

So far, we know that the equation of the axis of symmetry is x=- 1. Thus, the vertex lies at (- 1,k). To find the y-coordinate k, we will substitute - 1 for x in the given function.

y=- 3x^2-6x-8

Substitute

x= - 1

y=- 3( - 1)^2-6( - 1)-8

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Simplify right-hand side

CalcPow

Calculate power

y=- 3(1)-6(- 1)-8

IdPropMult

Identity Property of Multiplication

y=- 3-6(- 1)-8

MultNegNegOnePar

- a(- b)=a* b

y=- 3+6-8

AddSubTerms

Add and subtract terms

y=- 5

Therefore, the (h,k) coordinate pair of the vertex is (- 1,- 5). Moreover, since we already know that a= - 3, we can rewrite the given function in vertex form. y= - 3 (x-(- 1))^2+(- 5) ⇔ y=- 3(x+1)^2-5