Exercise 11 Page 549

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) = 1x^2 - 4x + 5 ⇕ f(x)=1x^2+( - 4)x+ 5We can see that a = 1, b = - 4, and c = 5. Since the y-intercept is given by the value of c, we know that the y-intercept is 5. Let's now substitute a=1 and b=- 4 into - b2a to find the axis of symmetry.

x = - b/2a

SubstituteII

a= 1, b= - 4

x = - - 4/2(1)

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Simplify

IdPropMult

Identity Property of Multiplication

x = - - 4/2

RemoveNegFracAndNum

- - a/b= a/b

x = 4/2

CalcQuot

Calculate quotient

x = 2

The equation of the axis of symmetry is x=2.
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=2. Therefore, the x-coordinate of the vertex is 2 and the y-coordinate is f(2). To find this value, substitute our x-coordinate for x in the given equation.

y=x^2-4x+5

Substitute

x= 2

y=( 2)^2-4( 2)+5

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Simplify right-hand side

CalcPow

Calculate power

y=4-4(2)+5

MultNegPos

(- a)b = - ab

y=4-8+5