Exercise 41 Page 550

Consider the general expression of a quadratic function, y=ax^2+ bx+ c, where a ≠ 0. Let's note three things we can learn from this equation.

The y-intercept is c.
The equation of the axis of symmetry is x=- b2a.
The x-coordinate of the vertex is - b2a.

We will start by identifying the values of a, b, and c. f(x)=- 3x^2+6x-18 ⇕ f(x)=(-3)x^2+ 6x+( - 18)We can see that a = - 3, b = 6, and c = - 18. Since the y-intercept is given by the value of c, we know that the y-intercept is - 18. Let's now substitute a=- 3 and b=6 into - b2a to find the axis of symmetry.

x = - b/2a

SubstituteII

a= - 3, b= 6

x = - 6/2(- 3)

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Simplify

MultPosNeg

a(- b)=- a * b

x = - 6/- 6

RemoveNegFracAndDenom

- a/- b= a/b

x = 6/6

QuotOne

a/a=1

x = 1

The equation of the axis of symmetry is x=1.
To find the vertex of the parabola, we will need to think of y as a function of x, y=f(x). We can write the expression for the vertex by stating the x- and y-coordinates in terms of a and b. Vertex: ( - b/2a, f(- b/2a ) ) When determining the axis of symmetry, we found that - b2a=1. Therefore, the x-coordinate of the vertex is 1 and the y-coordinate is f(1). To find this value, substitute our x-coordinate for x in the given equation.

f(x) = - 3 x^2+6x-18

Substitute

x= 1

f( 1) = - 3( 1)^2+6( 1)-18

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Simplify right-hand side

CalcPow

Calculate power

f(1) = - 3(1)+6(1)-18

IdPropMult

Identity Property of Multiplication

f(1) = - 3 + 6 - 18