Exercise 26 Page 647

We want to write the given quadratic equation in standard form and identify the vertex, axis of symmetry, and the direction of opening of its parabola.

Standard Form

We will first express the equation in standard form,

x = a (y - k)^{2} + h,

where

a,

h,

and

k

are either positive or negative constants. To do so, we will first rearrange the equation so that the

x -

terms and the constant are on the right-hand side.

x - 6 y = y^{2} + 4 \Leftrightarrow x = y^{2} + 6 y + 4

Let's now simplify the right-hand side by completing the square. We have to add and subtract

(\frac{b}{2})^{2} .

In this case, we have that the linear coefficient

b

is

6 .

b = 6 \Rightarrow (\frac{b}{2})^{2} = (\frac{6}{2})^{2} = 3^{2}

Let's do it!

x = y^{2} + 6 y + 4

Add and subtract $3^{2}$

x = (y^{2} + 6 y + 3^{2}) + 4 - 3^{2}

▼

a^{2} + 2 a b + b^{2} = (a + b)^{2}

SplitIntoFactors

Split into factors

x = (y^{2} + 2 y (3) + 3^{2}) + 4 - 3^{2}

FacPosPerfectSquare

$a^{2} + 2 a b + b^{2} = (a + b)^{2}$

x = (y + 3)^{2} + 4 - 3^{2}

▼

Solve for

x

CalcPow

Calculate power

x = (y + 3)^{2} + 4 - 9

SubTerm

Subtract term

x = (y + 3)^{2} - 5

In standard form the equation is written as

x = a (y - k)^{2} + h,

where

a,

h,

and

k

are either positive or negative constants.

x = (y + 3)^{2} - 5 ⇕ x = 1 (y - (- 3))^{2} + (- 5)

It is important to note that we do not need to graph the parabola to identify the desired information. Let's compare the general formula for the standard form to our equation.

General Formula : x = Equation : x = a (y - k)^{2} + - - h 1 (y - (- 3))^{2} + (- 5)

We can see that

a = 1,

k = - 3,

and

h = - 5 .

Vertex

The vertex of a quadratic function written in standard form is the point $(h, k) .$ For this exercise, we have $h = - 5$ and $k = - 3 .$ Therefore, the vertex of the given equation is $(- 5, - 3) .$

Axis of Symmetry

The axis of symmetry of the quadratic quadratic equation written in standard form is the horizontal line with equation $y = k .$ As we have already noticed, for our equation, this is $k = - 3 .$ Thus, the axis of symmetry is the line $y = - 3 .$

Direction of Opening

Recall that when the axis of symmetry is a horizontal line, if $a > 0,$ the parabola opens right. Conversely, if $a < 0,$ the parabola opens left. In the given equation, we have $a = 1,$ which is greater than $0 .$ Thus, the parabola opens right.