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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`(xh)^2=4a(yk)`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.`(xh)^2` `=` `4a(yk)` 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.`(xh)^2` `=` `4a``(yk)` Hence, the parabola is concave down.Therefore, it would make sense to say that the graph for `(xh)^2=4a(yk)` is 
Question 2 of 5
2. Question
Sketch the graph of `(x+2)^2=8(y1)`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(y1)` Since the coefficient of $$(y1)$$ 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(y1)` 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 directrixFinally, draw a parabola concave up 
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``divide4` `=` `2``divide4` 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 directrixFinally, draw a parabola concave down 
Question 4 of 5
4. Question
Sketch the graph of `(y+3)^2=2(x1)`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(x1)` Since the coefficient of $$(x1)$$ 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(x1)` 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 directrixFinally, draw a parabola concave right 
Question 5 of 5
5. Question
Sketch the graph of `(y2)^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.`(y2)^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$${(y\color{#007ddc}{k})}^2$$ `=` $$4\color{#9a00c7}{a}(x\color{#00880a}{h})$$ $${(y\color{#007ddc}{2})}^2$$ `=` $$(x\color{#00880a}{(3)})$$ `(y2)^2` `=` `(x+3)` The Vertex `(``h``,``k``)` can be read from the equation `(``3``,``2``)`Focal Length:`4a` `=` `1` `4a``divide4` `=` `1``divide4` 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 directrixFinally, draw a parabola concave left
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