Exercise 36 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. y=- 3x^2-6x+7 ⇕ y=(-3)x^2+( - 6)x+ 7We can see that a = - 3, b= - 6, and c = 7. Since the y-intercept is given by the value of c, we know that the y-intercept is 7. 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)

▼

Simplify

MultPosNeg

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.

y = - 3x^2-6x+7

Substitute

x= - 1

f( -1) = - 3( - 1)^2-6( - 1)+7

▼

Simplify right-hand side

CalcPow

Calculate power

f(-1) = - 3(1)-6(- 1)+7

MultNegNegOnePar

- a(- b)=a* b

f(-1) = - 3(1)+6+7