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Pearson Algebra 2 Common Core, 2011

PA

Pearson Algebra 2 Common Core, 2011 View details

2. Standard Form of a Quadratic Function

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Exercise 43 Page 207

Practice makes perfect

a The function that models the height of a projectile is

h (t) = 64 t - 16 t^{2} .

We can see that the coefficients of this quadratic function are

a = - 16,

b = 64,

and

c = 0 .

Since

a

is a negative number, the parabola opens downward and its maximum point is the vertex. Let's begin by finding the

x -

x = - \frac{b}{2 a}

Let's substitute

- 16

for

a

and

64

for

b .

t = - \frac{b}{2 a}

$a = - 16$ , $b = 64$

t = - \frac{6 4}{2 ( - 1 6 )}

▼

Simplify right-hand side

$a (- b) = - a \cdot b$

t = - \frac{6 4}{- 3 2}

Calculate quotient

t = - (- 2)

$- (- a) = a$

t = 2

To find the

y -

coordinate, we will substitute

2

for

t

into the the function

h (t) .

h (t) = 64 t - 16 t^{2}

$t = 2$

h (2) = 64 (2) - 16 (2)^{2}

▼

Simplify left-hand side

Calculate power

h (2) = 64 (2) - 16 (4)

Multiply

h (2) = 128 - 64

Subtract term

h (2) = 64

The projectile represented by the equation reached a maximum height of

64 ft

after

2

seconds. For the second projectile, we are given a table of values where we will highlight the rows with equal heights.

Time $(t)$	Height $(h)$
$0.5$	$20$
$1$	$32$
$1.5$	$36$
$2$	$32$

From the table above, the vertex must be the midpoint between $t = 1$ and $t = 2 .$ Therefore, the $x -$ coordinate of the vertex is $t = 1.5$ and its $y -$ coordinate is $36 .$ Let's plot the given data and trace the corresponding parabola.

From the graph, we conclude that the maximum height reached by the projectile represented by the table is $36 ft$ after $1.5$ seconds. The greatest height that the projectile represented by a function can reach is equal to $64 .$ This means the projectile represented by a function goes higher by $64 - 36 = 28 ft .$

b To determine when the projectile represented by the equation was at a height of

16 ft,

we must substitute

16

for the height in the equation

h (t) = 64 t - 16 t^{2}

and solve it for

t .

h (t) = 64 t - 16 t^{2}

$h (t) = 16$

16 = 64 t - 16 t^{2}

▼

Simplify

$LHS - 16 = RHS - 16$

0 = 64 t - 16 t^{2} - 16

CommutativePropAdd

Commutative Property of Addition

0 = - 16 t^{2} + 64 t - 16

$LHS / (- 16) = RHS / (- 16)$

0 = t^{2} - 4 t + 1

Rearrange equation

t^{2} - 4 t + 1 = 0

Next, we will use the Quadratic Formula with

a = 1,

b = - 4,

and

c = 1 .

t_{1, 2} = \frac{- b \pm b ^{2} - 4 a c}{2 a}

SubstituteValues

Substitute values

t_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( 1 ) ( 1 )}{2 ( 1 )}

▼

Simplify right-hand side

Multiply

t_{1, 2} = \frac{4 \pm 1 6 - 4}{2}

Subtract terms

t_{1, 2} = \frac{4 \pm 1 2}{2}

Calculate root

t_{1, 2} = \frac{4 \pm 3 . 4 6}{2}

Let's simplify

t_{1}

and

t_{2} .

$t_{1}$	$t_{2}$
$\frac{4 - 3 . 4 6}{2}$	$\frac{4 + 3 . 4 6}{2}$
$\frac{0 . 5 4}{2}$	$\frac{7 . 4 6}{2}$
$0.27$	$3.73$

Therefore, the projectile represented by the equation was at a height of

16 ft

at

t = 0.27

seconds and at

t = 3.73

seconds. For the projectile represented by the table, we must first find its equation. Let's write it in vertex form.

h_{2} (t) = a (t - h)^{2} + k

In this form, the vertex is

(h, k) .

From Part A we know the vertex is

(1.5, 36) .

h_{2} (t) = a (t - 1.5)^{2} + 36

To find the value of

a,

we will use the point

(0.5, 20)

from the table. Therefore,

h = 20

when

t = 0.5 .

h_{2} (t) = a (t - 1.5)^{2} + 36

$t = 0.5$

h_{2} (0.5) = a (0.5 - 1.5)^{2} + 36

$h (0.5) = 20$

20 = a (0.5 - 1.5)^{2} + 36

▼

Solve for

a

Subtract terms

20 = a (- 1)^{2} + 36

Calculate power

20 = a (1) + 36

$a \cdot 1 = a$

20 = a + 36

$LHS - 36 = RHS - 36$

- 16 = a

Rearrange equation

a = - 16

We can now write the equation that models the height of the projectile represented by the table.

h_{2} (t) = - 16 (t - 1.5)^{2} + 36

As before, we must substitute

16

for the height in the above equation.

h_{2} (t) = - 16 (t - 1.5)^{2} + 36

$h_{2} (t) = 16$

16 = - 16 (t - 1.5)^{2} + 36

▼

Solve for

t

$LHS - 36 = RHS - 36$

- 20 = - 16 (t - 1.5)^{2}

$LHS / (- 16) = RHS / (- 16)$

\frac{2 0}{1 6} = (t - 1.5)^{2}

$\frac{a}{b} = \frac{a / 4}{b / 4}$

\frac{5}{4} = (t - 1.5)^{2}

$LHS = RHS$

\frac{5}{4} = (t - 1.5)^{2}

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

\frac{5}{4} = (t - 1.5)^{2}

Calculate root

\frac{2 . 2 4}{2} = (t - 1.5)^{2}

Calculate quotient

1.12 = (t - 1.5)^{2}

$a^{2} = ∣ a ∣$

1.12 = ∣ t - 1.5 ∣

Rearrange equation

∣ t - 1.5 ∣ = 1.12

We can divide the final equation into two values, one with the negative sign and the other with the positive sign.

$∣ t - 1.5 ∣ = 1.12$
$t_{1} - 1.5 = - 1.12$	$t_{2} - 1.5 = 1.12$
$t_{1} = 0.38$	$t_{2} = 2.62$

In conclusion, the projectile represented by the table was at a height of $16 ft$ at $t = 0.38$ seconds and at $t = 2.62$ seconds.

Subchapter links

Lesson Check p.206

Practice and Problem-Solving Exercises p.206-208

Standardized Test Prep p.208

Mixed Review p.208