To determine To find: An equation of the form A x ² + D x + E y + F = 0 ; A ≠ 0 a. Is a parabola if E ≠ 0 . Answer a. This is the equation of the parabola whose vertex is ( − D 2 A , D 2 − 4 F A 4 A E ) and the axis of symmetry is parallel to y -axis . Explanation Given: An equation of the form A x ² + D x + E y + F = 0 ; A ≠ 0 Formula used: Vertex Focus Directrix Equation ( h , k ) ( h + a , k ) x = h − a ( y − k ) ² = 4 a ( x − h ) x = − b ± b 2 − 4 a c 2 a Calculation: a. If E ≠ 0 , A x ² + D x + E y + F = 0 A x ² + D x = − E y − F Adding D 2 4 A both sides. A ( x 2 + D A x + D 2 4 A ² ) = − E y − F + D 2 4 A A ( x + D 2 A ) 2 = 1 A ( − E y − F + D 2 4 A ) ( x + D 2 A ) 2 = − E A ( y + F E − D 2 4 A E ) ( x + D 2 A ) 2 = − E A ( y − D 2 − 4 F A 4 A E ) This is the equation of the parabola whose vertex is ( − D 2 A , D 2 − 4 F A 4 A E ) and the axis of symmetry is parallel to y -axis . To determine To find: An equation of the form A x ² + D x + E y + F = 0 ; A ≠ 0 b. Is a horizontal line if E = 0 and D ² − 4 A F = 0 . Answer b. It is a quadratic equation; x = − D ± D 2 − 4 A F 2 A if E ² = 4 A F , then x = − D / 2 A is a single horizontal line. Explanation Given: An equation of the form A x ² + D x + E y + F = 0 ; A ≠ 0 Formula used: Vertex Focus Directrix Equation ( h , k ) ( h + a , k ) x = h − a ( y − k ) ² = 4 a ( x − h ) x = − b ± b 2 − 4 a c 2 a Calculation: b. If E = 0 . A x ² + D x + E y + F = 0 A x ² + D x + F = 0 It is a quadratic equation; x = − D ± D 2 − 4 A F 2 A if E ² = 4 A F , then x = − D / 2 A is a single horizontal line. To determine To find: An equation of the form A x ² + D x + E y + F = 0 ; A ≠ 0 c. Is a horizontal line if E = 0 and D ² − 4 A F > 0 Answer c. It is a quadratic equation; x = − D ± D 2 − 4 A F 2 A if E ² > 4 A F , then x = − D + D ² − 4 A F 2 C , x = − D − D ² − 4 A F 2 C are two horizontal lines. Explanation Given: An equation of the form A x ² + D x + E y + F = 0 ; A ≠ 0 Formula used: Vertex Focus Directrix Equation ( h , k ) ( h + a , k ) x = h − a ( y − k ) ² = 4 a ( x − h ) x = − b ± b 2 − 4 a c 2 a Calculation: c. If E = 0 A x ² + D x + E y + F = 0 A x ² + D x + F = 0 It is a quadratic equation; x = − D ± D 2 − 4 A F 2 A if E ² > 4 A F , then x = − D + D ² − 4 A F 2 C , x = − D − D ² − 4 A F 2 C are two horizontal lines. To determine To find: An equation of the form A x ² + D x + E y + F = 0 ; A ≠ 0 d. Contains no points if E = 0 and D ² − 4 A F < 0 Answer d. It is a quadratic equation; x = − D ± D 2 − 4 A F 2 A if E ² < 4 A F , then there is no real solution for x and hence, the graph contains no points. Explanation Given: An equation of the form A x ² + D x + E y + F = 0 ; A ≠ 0 Formula used: Vertex Focus Directrix Equation ( h , k ) ( h + a , k ) x = h − a ( y − k ) ² = 4 a ( x − h ) x = − b ± b 2 − 4 a c 2 a Calculation: d. If E = 0 , then A x ² + D x + E y + F = 0 A x ² + D x + F = 0 It is a quadratic equation; x = − D ± D 2 − 4 A F 2 A if E ² < 4 A F , then there is no real solution for x and hence, the graph contains no points.

Precalculus Enhanced with Graphing Utilities

6th Edition

ISBN: 9780321795465

Author: Michael Sullivan, Michael III Sullivan

Publisher: PEARSON

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Concept explainers

Maximization

Have you ever seen a bridge in the shape of a parabola? Where is its maximum on the bridge? The maximum of that bridge is the highest point on that bridge. The process of finding the maximum of the bridge here is called the maximization.

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Textbook Question

Chapter 10.2, Problem 78AYU

Show that the graph of an equation of the form $A x^{2} + D x + E y + F = 0$ $A \neq 0$

(a) Is a parabola if $E \neq 0$ .

(b) Is a vertical line if $E = 0$ and $D^{2} - 4 A F = 0$ .

(d) Contains no points if $E = 0$ and $D^{2} - 4 A F < 0$ .

Expert Solution

To determine

To find: An equation of the form $A x ² + D x + E y + F = 0$ ; $A \neq 0$

a. Is a parabola if $E \neq 0$ .

Answer to Problem 78AYU

a. This is the equation of the parabola whose vertex is $(- \frac{D}{2 A}, \frac{D^{2} - 4 F A}{4 A E})$ and the axis of symmetry is parallel to $y -axis$ .

Explanation of Solution

Given:

An equation of the form $A x ² + D x + E y + F = 0$ ; $A \neq 0$

Formula used:

Vertex	Focus	Directrix	Equation
$(h, k)$	$(h + a, k)$	$x = h - a$	$(y - k) ² = 4 a (x - h)$

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

Calculation:

a. If $E \neq 0$ ,

$A x ² + D x + E y + F = 0$

$A x ² + D x = - E y - F$

Adding $\frac{D^{2}}{4 A}$ both sides.

$A (x^{2} + \frac{D}{A} x + \frac{D^{2}}{4 A ²}) = - E y - F + \frac{D^{2}}{4 A}$

$A {(x + \frac{D}{2 A})}^{2} = \frac{1}{A} (- E y - F + \frac{D^{2}}{4 A})$

${(x + \frac{D}{2 A})}^{2} = - \frac{E}{A} (y + \frac{F}{E} - \frac{D^{2}}{4 A E})$

${(x + \frac{D}{2 A})}^{2} = - \frac{E}{A} (y - \frac{D^{2} - 4 F A}{4 A E})$

This is the equation of the parabola whose vertex is $(- \frac{D}{2 A}, \frac{D^{2} - 4 F A}{4 A E})$ and the axis of symmetry is parallel to $y -axis$ .

Expert Solution

To determine

To find: An equation of the form $A x ² + D x + E y + F = 0$ ; $A \neq 0$

b. Is a horizontal line if $E = 0$ and $D ² - 4 A F = 0$ .

Answer to Problem 78AYU

b. It is a quadratic equation; $x = \frac{- D \pm \sqrt{D^{2} - 4 A F}}{2 A}$ if $E ² = 4 A F$ , then $x = - D / 2 A$ is a single horizontal line.

Explanation of Solution

Given:

An equation of the form $A x ² + D x + E y + F = 0$ ; $A \neq 0$

Formula used:

Vertex	Focus	Directrix	Equation
$(h, k)$	$(h + a, k)$	$x = h - a$	$(y - k) ² = 4 a (x - h)$

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

Calculation:

b. If $E = 0$ .

$A x ² + D x + E y + F = 0$

$A x ² + D x + F = 0$

It is a quadratic equation; $x = \frac{- D \pm \sqrt{D^{2} - 4 A F}}{2 A}$ if $E ² = 4 A F$ , then $x = - D / 2 A$ is a single horizontal line.

Expert Solution

To determine

To find: An equation of the form $A x ² + D x + E y + F = 0$ ; $A \neq 0$

c. Is a horizontal line if $E = 0$ and $D ² - 4 A F > 0$

Answer to Problem 78AYU

c. It is a quadratic equation; $x = \frac{- D \pm \sqrt{D^{2} - 4 A F}}{2 A}$ if $E ² > 4 A F$ , then $x = \frac{- D + \sqrt{D ² - 4 A F}}{2 C}$ , $x = \frac{- D - \sqrt{D ² - 4 A F}}{2 C}$ are two horizontal lines.

Explanation of Solution

Given:

An equation of the form $A x ² + D x + E y + F = 0$ ; $A \neq 0$

Formula used:

Vertex	Focus	Directrix	Equation
$(h, k)$	$(h + a, k)$	$x = h - a$	$(y - k) ² = 4 a (x - h)$

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

Calculation:

c. If $E = 0$

$A x ² + D x + E y + F = 0$

$A x ² + D x + F = 0$

It is a quadratic equation; $x = \frac{- D \pm \sqrt{D^{2} - 4 A F}}{2 A}$ if $E ² > 4 A F$ , then $x = \frac{- D + \sqrt{D ² - 4 A F}}{2 C}$ , $x = \frac{- D - \sqrt{D ² - 4 A F}}{2 C}$ are two horizontal lines.

Expert Solution

To determine

To find: An equation of the form $A x ² + D x + E y + F = 0$ ; $A \neq 0$

d. Contains no points if $E = 0$ and $D ² - 4 A F < 0$

Answer to Problem 78AYU

d. It is a quadratic equation; $x = \frac{- D \pm \sqrt{D^{2} - 4 A F}}{2 A}$ if $E ² < 4 A F$ , then there is no real solution for $x$ and hence, the graph contains no points.

Explanation of Solution

Given:

An equation of the form $A x ² + D x + E y + F = 0$ ; $A \neq 0$

Formula used:

Vertex	Focus	Directrix	Equation
$(h, k)$	$(h + a, k)$	$x = h - a$	$(y - k) ² = 4 a (x - h)$

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

Calculation:

d. If $E = 0$ , then

$A x ² + D x + E y + F = 0$

$A x ² + D x + F = 0$

It is a quadratic equation; $x = \frac{- D \pm \sqrt{D^{2} - 4 A F}}{2 A}$ if $E ² < 4 A F$ , then there is no real solution for $x$ and hence, the graph contains no points.

Chapter 10.2, Problem 77AYU

Chapter 10.2, Problem 79AYU

Chapter 10 Solutions

Precalculus Enhanced with Graphing Utilities

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