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Core Connections Algebra 1, 2013 View details

3. Section 10.3

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Exercise 139 Page 515

a We are given a quadratic function written in standard form. To determine the number of the

x -

intercepts we will rewrite the function in graphing form. We can do that by completing the square.

y = 6 x^{2} + 7 x - 20

Let's factor out

6

from the right-hand side of the rule, so the coefficient of

x^{2}

will be

1 .

y = 6 x^{2} + 7 x - 20

NumberToDenomMultFrac

$a = 6 \cdot \frac{a}{6}$

y = 6 x^{2} + 6 \cdot \frac{7}{6} x - 6 \cdot \frac{2 0}{6}

FactorOut

Factor out $6$

y = 6 (x^{2} + \frac{7}{6} x - \frac{2 0}{6})

ReduceFrac

$\frac{a}{b} = \frac{a / 2}{b / 2}$

y = 6 (x^{2} + \frac{7}{6} x - \frac{1 0}{3})

In a quadratic expression,

b

is the linear coefficient. For the function above we have that

b = \frac{7}{6} .

Let's now calculate

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

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

Substitute

$b = \frac{7}{6}$

(\frac{7 / 6}{2})^{2}

▼

Simplify

DivFracD

$\frac{a / c}{b} = \frac{a}{b \cdot c}$

(\frac{7}{1 2})^{2}

PowQuot

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

\frac{7 ^{2}}{1 2 ^{2}}

CalcPow

Calculate power

\frac{4 9}{1 4 4}

Next, we will add and subtract

(\frac{b}{2})^{2} = \frac{4 9}{1 4 4}

to our function. Then we will factor the obtained perfect square trinomial.

y = 6 (x^{2} + \frac{7}{6} x - \frac{1 0}{3})

CompleteSquare

Complete the square

y = 6 (x^{2} + \frac{7}{6} x + \frac{4 9}{1 4 4} - \frac{1 0}{3} - \frac{4 9}{1 4 4})

FacPosPerfectSquare

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

y = 6 ((x + \frac{7}{1 2})^{2} - \frac{1 0}{3} - \frac{4 9}{1 4 4})

▼

Simplify

ExpandFrac

$\frac{a}{b} = \frac{a \cdot 4 8}{b \cdot 4 8}$

y = 6 ((x + \frac{7}{1 2})^{2} - \frac{4 8 0}{1 4 4} - \frac{4 9}{1 4 4})

SubFrac

Subtract fractions

y = 6 ((x + \frac{7}{1 2})^{2} - \frac{5 2 9}{1 4 4})

Distr

Distribute $6$

y = 6 (x + \frac{7}{1 2})^{2} - 6 (\frac{5 2 9}{1 4 4})

CancelCommonFac

Cancel out common factors

y = 6 (x + \frac{7}{1 2})^{2} - \frac{5 2 9}{2 4}

Finally, we expressed our function in graphing form. For our function

a = 6,

which is greater than

0 .

Therefore, the parabola opens upwards. We want to find the number of

x -

intercepts. Notice that we can find it by determining whether there is any value of

x

for which the function is below the

x -

axis.

Let's analyze the rule that we found. It contains two parts: a square and a constant

(c) .

y = square 6 (x + \frac{7}{1 2})^{2} - \frac{5 2 9}{2 4} \leftarrow constant

Regardless of what number we substitute for

x,

the square part will always be non-negative Therefore, the least value of the square part is

0 .

It happens when

x

- \frac{7}{1 2} .

For all other values of

x,

the square part is positive. Thus, to see if there is an

x -

value for which the corresponding

y -

value is negative, we need to look at the sign of the constant

(c) .

Let's sum up what we have found.

If the constant is positive — there will be no $x -$ intercepts. This is because the square part is non-negative and adding a positive constant to it will always give us a positive value of $y .$ This means that the graph will be above the $x -$ axis.
If the constant is negative — there will be two $x -$ intercepts. This is because when the square part is equal to $0,$ the corresponding $y -$ value is the negative value of $c,$ which means that the graph will be below the $x -$ axis.
If the constant is equal to zero — the lowest point of the graph will lay on the $x -$ axis. Thus, there will be only one $x -$ intercept.

Notice that in our function, represented in the graphing form, the constant is equal to $- \frac{5 2 9}{2 4} .$ It is negative, so according to the above reasoning our function has two $x -$ intercepts.

b Just like in Part A, to determine the number of the

x -

intercepts we will rewrite the function in graphing form. This time, before we will try to complete the square, let's check if the function is not already a perfect square trinomial. To determine if an expression is a perfect square trinomial, we need to answer three questions.

Is the first term a perfect square?	$x^{2} = (x)^{2}$ $✓$
Is the last term a perfect square?	$16 = 4^{2}$ $✓$
Is the middle term $twice$ the product of $4$ and $x$ ?	$8 x = 2 \cdot 4 \cdot x$ $✓$

As we can see, the answer to all three questions is yes! Therefore, we can write the trinomial as the square of a binomial. Note there is a subtraction sign in the middle.

y = x^{2} - 8 x + 16 \Leftrightarrow y = (x - 4)^{2}

We expressed our function in graphing form.

y = square (x - 4)^{2} + 0 \leftarrow constant

As we already discussed in Part A, when the constant is equal zero, the graph of the function has exactly one

x -

intercept. Thus, the graph of the given rule has one

x -

intercept.

c We have a quadratic function written in standard form. To determine the number of

x -

intercepts we will rewrite the function in graphing form. We can do that by completing the square.

y = 2 x^{2} + x + 3

Now let's factor out

2

from the right-hand side of the rule, so the coefficient of

x^{2}

will be

1 .

y = 2 x^{2} + x + 3

NumberToDenomMultFrac

$a = 2 \cdot \frac{a}{2}$

y = 2 x^{2} + 2 \cdot \frac{1}{2} x + 2 \cdot \frac{3}{2}

FactorOut

Factor out $2$

y = 2 (x^{2} + \frac{1}{2} x - \frac{3}{2})

In a quadratic expression

b

is the linear coefficient. For the function above, we have that

b = \frac{1}{2} .

Let's now calculate

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

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

Substitute

$b = \frac{1}{2}$

(\frac{1 / 2}{2})^{2}

▼

Simplify

DivFracD

$\frac{a / c}{b} = \frac{a}{b \cdot c}$

(\frac{1}{4})^{2}

PowQuot

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

\frac{1 ^{2}}{4 ^{2}}

CalcPow

Calculate power

\frac{1}{1 6}

Next, we will add and subtract

(\frac{b}{2})^{2} = \frac{1}{1 6}

to our function. Then we will factor the obtained perfect square trinomial.

y = 2 (x^{2} + \frac{1}{2} x + \frac{3}{2})

CompleteSquare

Complete the square

y = 2 (x^{2} + \frac{1}{2} x + \frac{1}{1 6} + \frac{3}{2} - \frac{1}{1 6})

FacPosPerfectSquare

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

y = 2 ((x + \frac{1}{4})^{2} + \frac{3}{2} - \frac{1}{1 6})

▼

Simplify

ExpandFrac

$\frac{a}{b} = \frac{a \cdot 8}{b \cdot 8}$

y = 2 ((x + \frac{1}{4})^{2} + \frac{2 4}{1 6} - \frac{1}{1 6})

SubFrac

Subtract fractions

y = 2 ((x + \frac{1}{4})^{2} - \frac{2 3}{1 6})

Distr

Distribute $2$

y = 2 (x + \frac{1}{4})^{2} + 2 (\frac{2 3}{1 6})

CancelCommonFac

Cancel out common factors

y = 2 (x + \frac{1}{4})^{2} + \frac{2 3}{8}

Finally, we expressed our function in graphing form.

y = square 2 (x + \frac{1}{4})^{2} + \frac{2 3}{8} \leftarrow constant

Notice that in our function, represented in the graphing form, the constant is positive. Thus, as we already found in Part A, the graph of our function is above the