Graphing Parabolas (Focus and Directrix) 2

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Question 1 of 5

1. Question
Which of the following will most likely be the graph of

${(x - h)}^{2} = - 4 a (y - k)$ $(x-h)^2=-4a(y-k)$
- 1.
- 2.
- 3.
- 4.
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In a quadratic equation, if the coefficient of the $y$ $y$ term is positive, the parabola will be concave up. If it is negative, the parabola will be concave down

Notice that the equation given is in vertex form.

${(x - h)}^{2}$ $(x-h)^2$ $=$ $=$ $- 4 a (y - k)$ $-4a(y-k)$

This means that the vertex is at $(h, k)$ $(h,k)$ .

Also, take note of the following components:

Focus $(h, a + k)$ $(h,a+k)$

this point lies in the parabola’s axis of symmetry and is at equal distance with all points on the parabola

Directrix $y = - a$ $y=-a$

this line lies in the opposite direction of the parabola and is at equal distance with all points on the parabola

Next, check the sign of the coefficient of the $y$ $y$ term.

${(x - h)}^{2}$ $(x-h)^2$ $=$ $=$ $- 4 a$ $-4a$ $(y - k)$ $(y-k)$

Hence, the parabola is concave down.

Therefore, it would make sense to say that the graph for ${(x - h)}^{2} = - 4 a (y - k)$ $(x-h)^2=-4a(y-k)$ is

Question 2 of 5

Sketch the graph of

{(x + 2)}^{2} = 8 (y - 1)

$(x+2)^2=8(y-1)$

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Standard Form (Concave Up)

{(x - h)}^{2} = 4 a (y - k)

${(x-\color{#00880a}{h})}^2=4\color{#9a00c7}{a}(y-\color{#007ddc}{k})$

Focus

(

$($ $h$ $h$

,

$,$ $a$ $a$

+

$+$ $k$ $k$

)

$)$

Standard Form (Concave Down)

{(x - h)}^{2} = - 4 a (y - k)

${(x-\color{#00880a}{h})}^2=-4\color{#9a00c7}{a}(y-\color{#007ddc}{k})$

Focus

(

$($ $h$ $h$

, -

$,-$ $a$ $a$

+

$+$ $k$ $k$

)

$)$

First, determine which standard form applies to the given equation.

{(x + 2)}^{2}

$(x+2)^2$

=

$=$

8 (y - 1)

$8(y-1)$

Since the coefficient of

(y - 1)

$(y-1)$ is positive, use

{(x - h)}^{2} = 4 a (y - k)

${(x-\color{#00880a}{h})}^2=4\color{#9a00c7}{a}(y-\color{#007ddc}{k})$

This also means that the parabola will be concave up

Find the focal length ( $a$ $a$ ) and vertex by comparing the given equation to the standard form

$(x-\color{#00880a}{h})^2$	$=$ $=$	$4\color{#9a00c7}{a}(y-\color{#007ddc}{k})$
$(x-\color{#00880a}{(-2)})^2$	$=$ $=$	$8(y-\color{#007ddc}{1})$
${(x + 2)}^{2}$ $(x+2)^2$	$=$ $=$	$8 (y - 1)$ $8(y-1)$

The Vertex

(

$($ $h$ $h$

,

$,$ $k$ $k$

)

$)$ can be read from the equation

(

$($ $- 2$ $-2$

,

$,$ $1$ $1$

)

$)$

Focal Length:

$4 a$ $4a$	$=$ $=$	$8$ $8$
$4 a$ $4a$ $\div 4$ $divide4$	$=$ $=$	$8$ $8$ $\div 4$ $divide4$	Divide both sides by $4$ $4$
$a$ $a$	$=$ $=$	$2$ $2$

Identify the focus and directrix.

Focus:

(

$($ $h$ $h$

,

$,$ $a$ $a$

+

$+$ $k$ $k$

)

$)$ becomes

(

$($ $- 2$ $-2$

,

$,$ $2$ $2$

+

$+$ $1$ $1$

)

$)$ or

(- 2, 3)

$(-2,3)$

Directrix:

y = -

$y=-$ $a$ $a$

+

$+$ $k$ $k$ becomes

y = -

$y=-$ $2$ $2$

+

$+$ $1$ $1$ or

y = - 1

$y=-1$

Start graphing the parabola by plotting the vertex, focus and directrix

Finally, draw a parabola concave up

Question 3 of 5

Sketch the graph of

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

$x^2=-2(y+5)$

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Standard Form (Concave Up)

{(x - h)}^{2} = 4 a (y - k)

${(x-\color{#00880a}{h})}^2=4\color{#9a00c7}{a}(y-\color{#007ddc}{k})$

Focus

(

$($ $h$ $h$

,

$,$ $a$ $a$

+

$+$ $k$ $k$

)

$)$

Standard Form (Concave Down)

{(x - h)}^{2} = - 4 a (y - k)

${(x-\color{#00880a}{h})}^2=-4\color{#9a00c7}{a}(y-\color{#007ddc}{k})$

Focus

(

$($ $h$ $h$

, -

$,-$ $a$ $a$

+

$+$ $k$ $k$

)

$)$

First, determine which standard form applies to the given equation.

x^{2}

$x^2$

=

$=$

- 2 (y + 5)

$-2(y+5)$

Since the coefficient of

(y + 5)

$(y+5)$ is negative, use

{(x - h)}^{2} = - 4 a (y - k)

${(x-\color{#00880a}{h})}^2=-4\color{#9a00c7}{a}(y-\color{#007ddc}{k})$

This also means that the parabola will be concave down

Find the focal length ( $a$ $a$ ) and vertex by comparing the given equation to the standard form

$(x-\color{#00880a}{h})^2$	$=$ $=$	$-4\color{#9a00c7}{a}(y-\color{#007ddc}{k})$
$(x-\color{#00880a}{0})^2$	$=$ $=$	$-2(y-\color{#007ddc}{(-5)})$
$x^{2}$ $x^2$	$=$ $=$	$- 2 (y + 5)$ $-2(y+5)$

The Vertex

(

$($ $h$ $h$

,

$,$ $k$ $k$

)

$)$ can be read from the equation

(

$($ $0$ $0$

,

$,$ $- 5$ $-5$

)

$)$

Focal Length:

$- 4 a$ $-4a$	$=$ $=$	$- 2$ $-2$
$- 4 a$ $-4a$ $\div - 4$ $divide-4$	$=$ $=$	$- 2$ $-2$ $\div - 4$ $divide-4$	Divide both sides by $- 4$ $-4$

$a$ $a$	$=$ $=$	$\frac{1}{2}$ $1/2$

Identify the focus and directrix.

Focus:

(

$($ $h$ $h$

, -

$,-$ $a$ $a$

+

$+$ $k$ $k$

)

$)$ becomes

(

$($ $0$ $0$

, -

$,-$ $\frac{1}{2}$ $1/2$

+

$+$ $(- 5)$ $(-5)$

)

$)$ or

(0, - 5 \frac{1}{2})

$(0,-5 1/2)$

Directrix:

y =

$y=$ $a$ $a$

+

$+$ $k$ $k$ becomes

y =

$y=$ $\frac{1}{2}$ $1/2$

+

$+$ $(- 5)$ $(-5)$ or

y = - 4 \frac{1}{2}

$y=-4 1/2$

Start graphing the parabola by plotting the vertex, focus and directrix

Finally, draw a parabola concave down

Question 4 of 5

Sketch the graph of

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

$(y+3)^2=2(x-1)$

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Standard Form (Concave Right)

{(y - k)}^{2} = 4 a (x - h)

${(y-\color{#007ddc}{k})}^2=4\color{#9a00c7}{a}(x-\color{#00880a}{h})$

Focus

(

$($ $a$ $a$

+

$+$ $h$ $h$

,

$,$ $k$ $k$

)

$)$

Standard Form (Concave Left)

{(y - k)}^{2} = - 4 a (x - h)

${(y-\color{#007ddc}{k})}^2=-4\color{#9a00c7}{a}(x-\color{#00880a}{h})$

Focus

(-

$(-$ $a$ $a$

+

$+$ $h$ $h$

,

$,$ $k$ $k$

)

$)$

First, determine which standard form applies to the given equation.

{(y + 3)}^{2}

$(y+3)^2$

=

$=$

2 (x - 1)

$2(x-1)$

Since the coefficient of

(x - 1)

$(x-1)$ is positive, use

{(y - k)}^{2} = 4 a (x - h)

${(y-\color{#007ddc}{k})}^2=4\color{#9a00c7}{a}(x-\color{#00880a}{h})$

This also means that the parabola will be concave right

Find the focal length ( $a$ $a$ ) and vertex by comparing the given equation to the standard form

${(y-\color{#007ddc}{k})}^2$	$=$ $=$	$4\color{#9a00c7}{a}(x-\color{#00880a}{h})$
${(y-\color{#007ddc}{(-3)})}^2$	$=$ $=$	$2(x-\color{#00880a}{1})$
${(y + 3)}^{2}$ $(y+3)^2$	$=$ $=$	$2 (x - 1)$ $2(x-1)$

The Vertex

(

$($ $h$ $h$

,

$,$ $k$ $k$

)

$)$ can be read from the equation

(

$($ $1$ $1$

,

$,$ $- 3$ $-3$

)

$)$

Focal Length:

$4 a$ $4a$	$=$ $=$	$2$ $2$
$4 a$ $4a$ $\div 4$ $divide4$	$=$ $=$	$2$ $2$ $\div 4$ $divide4$	Divide both sides by $4$ $4$

$a$ $a$	$=$ $=$	$\frac{1}{2}$ $1/2$

Identify the focus and directrix.

Focus:

(

$($ $a$ $a$

+

$+$ $h$ $h$

,

$,$ $k$ $k$

)

$)$ becomes

(

$($ $\frac{1}{2}$ $1/2$

+

$+$ $1$ $1$

,

$,$ $- 3$ $-3$

)

$)$ or

(1 \frac{1}{2}, - 3)

$(1 1/2,-3)$

Directrix:

x = -

$x=-$ $a$ $a$

+

$+$ $h$ $h$ becomes

x = -

$x=-$ $\frac{1}{2}$ $1/2$

+

$+$ $1$ $1$ or

x = \frac{1}{2}

$x=1/2$

Start graphing the parabola by plotting the vertex, focus and directrix

Finally, draw a parabola concave right

Question 5 of 5

Sketch the graph of

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

$(y-2)^2=-(x+3)$

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Standard Form (Concave Right)

{(y - k)}^{2} = 4 a (x - h)

${(y-\color{#007ddc}{k})}^2=4\color{#9a00c7}{a}(x-\color{#00880a}{h})$

Focus

(

$($ $a$ $a$

+

$+$ $h$ $h$

,

$,$ $k$ $k$

)

$)$

Standard Form (Concave Left)

{(y - k)}^{2} = - 4 a (x - h)

${(y-\color{#007ddc}{k})}^2=-4\color{#9a00c7}{a}(x-\color{#00880a}{h})$

Focus

(-

$(-$ $a$ $a$

+

$+$ $h$ $h$

,

$,$ $k$ $k$

)

$)$

First, determine which standard form applies to the given equation.

{(y - 2)}^{2}

$(y-2)^2$

=

$=$

- (x + 3)

$-(x+3)$

Since the coefficient of

(x + 3)

$(x+3)$ is negative, use

{(y - k)}^{2} = - 4 a (x - h)

${(y-\color{#007ddc}{k})}^2=-4\color{#9a00c7}{a}(x-\color{#00880a}{h})$

This also means that the parabola will be concave left

Find the focal length ( $a$ $a$ ) and vertex by comparing the given equation to the standard form

${(y-\color{#007ddc}{k})}^2$	$=$ $=$	$-4\color{#9a00c7}{a}(x-\color{#00880a}{h})$
${(y-\color{#007ddc}{2})}^2$	$=$ $=$	$-(x-\color{#00880a}{(-3)})$
${(y - 2)}^{2}$ $(y-2)^2$	$=$ $=$	$- (x + 3)$ $-(x+3)$

The Vertex

(

$($ $h$ $h$

,

$,$ $k$ $k$

)

$)$ can be read from the equation

(

$($ $- 3$ $-3$

,

$,$ $2$ $2$

)

$)$

Focal Length:

$- 4 a$ $-4a$	$=$ $=$	$- 1$ $-1$
$- 4 a$ $-4a$ $\div - 4$ $divide-4$	$=$ $=$	$- 1$ $-1$ $\div - 4$ $divide-4$	Divide both sides by $- 4$ $-4$

$a$ $a$	$=$ $=$	$\frac{1}{4}$ $1/4$

Identify the focus and directrix.

Focus:

(-

$(-$ $a$ $a$

+

$+$ $h$ $h$

,

$,$ $k$ $k$

)

$)$ becomes

(-

$(-$ $\frac{1}{4}$ $1/4$

+

$+$ $(- 3)$ $(-3)$

,

$,$ $2$ $2$

)

$)$ or

(- 3 \frac{1}{4}, 2)

$(-3 1/4,2)$

Directrix:

x =

$x=$ $a$ $a$

+

$+$ $h$ $h$ becomes

x =

$x=$ $\frac{1}{4}$ $1/4$

+

$+$ $(- 3)$ $(-3)$ or

x = - 2 \frac{3}{4}

$x=-2 3/4$

Start graphing the parabola by plotting the vertex, focus and directrix

Finally, draw a parabola concave left

Graphing Parabolas (Focus and Directrix) 2

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Quiz summary

Information

1. Question

Hint

Great Work!

2. Question

Hint

Correct!

Standard Form (Concave Up)

Focus

Standard Form (Concave Down)

Focus

3. Question

Hint

Keep Going!

Standard Form (Concave Up)

Focus

Standard Form (Concave Down)

Focus

4. Question

Hint

Fantastic!

Standard Form (Concave Right)

Focus

Standard Form (Concave Left)

Focus

5. Question

Hint

Excellent!

Standard Form (Concave Right)

Focus

Standard Form (Concave Left)

Focus

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