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Graph Parabolas (Focus and Directrix)>
Graph Parabolas (Focus and Directrix) 2Graph 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=-4a(y-k)`Hint
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In a quadratic equation, if the coefficient of the `y` term is positive, the parabola will be concave up. If it is negative, the parabola will be concave downNotice that the equation given is in vertex form.`(x-h)^2` `=` `-4a(y-k)` This means that the vertex is at `(h,k)`.Also, take note of the following components:Focus `(h,a+k)`this point lies in the parabola’s axis of symmetry and is at equal distance with all points on the parabolaDirectrix `y=-a`this line lies in the opposite direction of the parabola and is at equal distance with all points on the parabolaNext, check the sign of the coefficient of the `y` term.`(x-h)^2` `=` `-4a``(y-k)` Hence, the parabola is concave down.Therefore, it would make sense to say that the graph for `(x-h)^2=-4a(y-k)` is
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Question 2 of 5
2. Question
Sketch the graph of `(x+2)^2=8(y-1)`Hint
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Standard Form (Concave Up)
$${(x-\color{#00880a}{h})}^2=4\color{#9a00c7}{a}(y-\color{#007ddc}{k})$$Focus
`(``h``,``a``+``k``)`Standard Form (Concave Down)
$${(x-\color{#00880a}{h})}^2=-4\color{#9a00c7}{a}(y-\color{#007ddc}{k})$$Focus
`(``h``,-``a``+``k``)`First, determine which standard form applies to the given equation.`(x+2)^2` `=` `8(y-1)` Since the coefficient of $$(y-1)$$ is positive, use $${(x-\color{#00880a}{h})}^2=4\color{#9a00c7}{a}(y-\color{#007ddc}{k})$$This also means that the parabola will be concave upFind the focal length (`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` `=` `8(y-1)` The Vertex `(``h``,``k``)` can be read from the equation `(``-2``,``1``)`Focal Length:`4a` `=` `8` `4a``divide4` `=` `8``divide4` Divide both sides by `4` `a` `=` `2` Identify the focus and directrix.Focus:`(``h``,``a``+``k``)` becomes `(``-2``,``2``+``1``)` or `(-2,3)`Directrix:`y=-``a``+``k` becomes `y=-``2``+``1` or `y=-1`Start graphing the parabola by plotting the vertex, focus and directrix
Finally, draw a parabola concave up
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Question 3 of 5
3. Question
Sketch the graph of `x^2=-2(y+5)`Hint
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Standard Form (Concave Up)
$${(x-\color{#00880a}{h})}^2=4\color{#9a00c7}{a}(y-\color{#007ddc}{k})$$Focus
`(``h``,``a``+``k``)`Standard Form (Concave Down)
$${(x-\color{#00880a}{h})}^2=-4\color{#9a00c7}{a}(y-\color{#007ddc}{k})$$Focus
`(``h``,-``a``+``k``)`First, determine which standard form applies to the given equation.`x^2` `=` `-2(y+5)` Since the coefficient of $$(y+5)$$ is negative, use $${(x-\color{#00880a}{h})}^2=-4\color{#9a00c7}{a}(y-\color{#007ddc}{k})$$This also means that the parabola will be concave downFind the focal length (`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` `=` `-2(y+5)` The Vertex `(``h``,``k``)` can be read from the equation `(``0``,``-5``)`Focal Length:`-4a` `=` `-2` `-4a``divide-4` `=` `-2``divide-4` Divide both sides by `-4` `a` `=` `1/2` Identify the focus and directrix.Focus:`(``h``,-``a``+``k``)` becomes `(``0``,-``1/2``+``(-5)``)` or `(0,-5 1/2)`Directrix:`y=``a``+``k` becomes `y=``1/2``+``(-5)` or `y=-4 1/2`Start graphing the parabola by plotting the vertex, focus and directrix
Finally, draw a parabola concave down
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Question 4 of 5
4. Question
Sketch the graph of `(y+3)^2=2(x-1)`Hint
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Standard Form (Concave Right)
$${(y-\color{#007ddc}{k})}^2=4\color{#9a00c7}{a}(x-\color{#00880a}{h})$$Focus
`(``a``+``h``,``k``)`Standard Form (Concave Left)
$${(y-\color{#007ddc}{k})}^2=-4\color{#9a00c7}{a}(x-\color{#00880a}{h})$$Focus
`(-``a``+``h``,``k``)`First, determine which standard form applies to the given equation.`(y+3)^2` `=` `2(x-1)` Since the coefficient of $$(x-1)$$ is positive, use $${(y-\color{#007ddc}{k})}^2=4\color{#9a00c7}{a}(x-\color{#00880a}{h})$$This also means that the parabola will be concave rightFind the focal length (`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` `=` `2(x-1)` The Vertex `(``h``,``k``)` can be read from the equation `(``1``,``-3``)`Focal Length:`4a` `=` `2` `4a``divide4` `=` `2``divide4` Divide both sides by `4` `a` `=` `1/2` Identify the focus and directrix.Focus:`(``a``+``h``,``k``)` becomes `(``1/2``+``1``,``-3``)` or `(1 1/2,-3)`Directrix:`x=-``a``+``h` becomes `x=-``1/2``+``1` or `x=1/2`Start graphing the parabola by plotting the vertex, focus and directrix
Finally, draw a parabola concave right
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Question 5 of 5
5. Question
Sketch the graph of `(y-2)^2=-(x+3)`Hint
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Standard Form (Concave Right)
$${(y-\color{#007ddc}{k})}^2=4\color{#9a00c7}{a}(x-\color{#00880a}{h})$$Focus
`(``a``+``h``,``k``)`Standard Form (Concave Left)
$${(y-\color{#007ddc}{k})}^2=-4\color{#9a00c7}{a}(x-\color{#00880a}{h})$$Focus
`(-``a``+``h``,``k``)`First, determine which standard form applies to the given equation.`(y-2)^2` `=` `-(x+3)` Since the coefficient of $$(x+3)$$ is negative, use $${(y-\color{#007ddc}{k})}^2=-4\color{#9a00c7}{a}(x-\color{#00880a}{h})$$This also means that the parabola will be concave leftFind the focal length (`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` `=` `-(x+3)` The Vertex `(``h``,``k``)` can be read from the equation `(``-3``,``2``)`Focal Length:`-4a` `=` `-1` `-4a``divide-4` `=` `-1``divide-4` Divide both sides by `-4` `a` `=` `1/4` Identify the focus and directrix.Focus:`(-``a``+``h``,``k``)` becomes `(-``1/4``+``(-3)``,``2``)` or `(-3 1/4,2)`Directrix:`x=``a``+``h` becomes `x=``1/4``+``(-3)` or `x=-2 3/4`Start graphing the parabola by plotting the vertex, focus and directrix
Finally, draw a parabola concave left
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