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Year 11>
Locus and The Parabola>
Graphing Parabolas Given in General Form>
Graphing Parabolas Given in General FormGraphing Parabolas Given in General Form
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Question 1 of 4
1. Question
Sketch the graph of `x^2+2x-4y-11=0`Hint
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Excellent!
Incorrect
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, transform the given equation into vertex form.`x^2+2x-4y-11` `=` `0` `x^2+2x` `=` `4y+11` Keep only `x` terms on the left side `x^2+2x` `+1` `=` `4y+11` `+1` Add `1` to both sides to complete the square `(x+1)^2` `=` `4y+12` Contract left side `(x+1)^2` `=` `4(y+3)` Factor right side Now, determine which standard form applies to the given equation.`(x+1)^2` `=` `4(y+3)` Since the coefficient of $$(y+3)$$ 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$$(x-\color{#00880a}{h})^2$$ `=` $$4\color{#9a00c7}{a}(y-\color{#007ddc}{k})$$ $$(x-\color{#00880a}{(-1)})^2$$ `=` $$4(y-\color{#007ddc}{(-3)})$$ `(x+1)^2` `=` `4(y+3)` The Vertex `(``h``,``k``)` can be read from the equation `(``-1``,``-3``)`Focal Length:`4a` `=` `4` `4a``divide4` `=` `4``divide4` Divide both sides by `4` `a` `=` `1` Identify the focus and directrix.Focus:`(``h``,``a``+``k``)` becomes `(``-1``,``1``+``(-3)``)` or `(-1,-2)`Directrix:`y=-``a``+``k` becomes `y=-``1``+``(-3)` or `y=-4`Start graphing the parabola by plotting the vertex, focus and directrix
Finally, draw a parabola concave up
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Question 2 of 4
2. Question
Sketch the graph of `x^2+4x+8y-4=0`Hint
Help VideoCorrect
Well Done!
Incorrect
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, transform the given equation into vertex form.`x^2+4x+8y-4` `=` `0` `x^2+4x` `=` `-8y+4` Keep only `x` terms on the left side `x^2+4x` `+4` `=` `-8y+4` `+4` Add `4` to both sides to complete the square `(x+2)^2` `=` `-8y+8` Contract left side `(x+2)^2` `=` `-8(y-1)` Factor right side Now, determine which standard form applies to the given equation.`(x+2)^2` `=` `-8(y-1)` Since the coefficient of $$(y-2)$$ 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$$(x-\color{#00880a}{h})^2$$ `=` $$-4\color{#9a00c7}{a}(y-\color{#007ddc}{k})$$ $$(x-\color{#00880a}{(-2)})^2$$ `=` $$-8(y-\color{#007ddc}{1})$$ `(x+2)^2` `=` `-8(y-1)` The Vertex `(``h``,``k``)` can be read from the equation `(``-2``,``1``)`Focal Length:`-4a` `=` `-8` `-4a``divide-4` `=` `-8``divide-4` Divide both sides by `-4` `a` `=` `2` Identify the focus and directrix.Focus:`(``h``,-``a``+``k``)` becomes `(``-2``,-``2``+``1``)` or `(-2,-1)`Directrix:`y=``a``+``k` becomes `y=``2``+``1` or `y=3`Start graphing the parabola by plotting the vertex, focus and directrix
Finally, draw a parabola concave down
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Question 3 of 4
3. Question
Sketch the graph of `y^2+2y-12x+13=0`Hint
Help VideoCorrect
Nice Job!
Incorrect
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, transform the given equation into vertex form.`y^2+2y-12x+13` `=` `0` `y^2+2y` `=` `12x-13` Keep only `y` terms on the left side `y^2+2y` `+1` `=` `12x-13` `+1` Add `1` to both sides to complete the square `(y+1)^2` `=` `12x-12` Contract left side `(y+1)^2` `=` `12(x-1)` Factor right side First, determine which standard form applies to the given equation.`(y+1)^2` `=` `12(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$${(y-\color{#007ddc}{k})}^2$$ `=` $$4\color{#9a00c7}{a}(x-\color{#00880a}{h})$$ $${(y-\color{#007ddc}{(-1)})}^2$$ `=` $$12(x-\color{#00880a}{1})$$ `(y+1)^2` `=` `12(x-1)` The Vertex `(``h``,``k``)` can be read from the equation `(``1``,``-1``)`Focal Length:`4a` `=` `12` `4a``divide4` `=` `12``divide4` Divide both sides by `4` `a` `=` `3` Identify the focus and directrix.Focus:`(``a``+``h``,``k``)` becomes `(``3``+``1``,``(-1)``)` or `(4,-1)`Directrix:`x=-``a``+``h` becomes `x=-``3``+``1` or `x=-2`Start graphing the parabola by plotting the vertex, focus and directrix
Finally, draw a parabola concave right
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Question 4 of 4
4. Question
Sketch the graph of `x=1/4(y^2-6y+13)`Hint
Help VideoCorrect
Fantastic!
Incorrect
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, transform the given equation into vertex form.`x` `=` `1/4(y^2-6y+13)` `x``times4` `=` `1/4(y^2-6y+13)``times4` Multiply both sides by `4` `4x` `=` `y^2-6y+13` `y^2-6y` `=` `4x-13` Keep only `y` terms on the left side `y^2-6y` `+9` `=` `4x-13` `+9` Complete the square by adding to both sides: `(6divide2)^2=9` `(y-3)^2` `=` `4x-4` Contract left side `(y-3)^2` `=` `4(x-1)` Factor right side First, determine which standard form applies to the given equation.`(y-3)^2` `=` `4(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$${(y-\color{#007ddc}{k})}^2$$ `=` $$4\color{#9a00c7}{a}(x-\color{#00880a}{h})$$ $${(y-\color{#007ddc}{3})}^2$$ `=` $$4(x-\color{#00880a}{1})$$ The Vertex `(``h``,``k``)` can be read from the equation `(``1``,``3``)`Focal Length:`4a` `=` `4` `4a``divide4` `=` `4``divide4` Divide both sides by `4` `a` `=` `1` Identify the focus and directrix.Focus:`(``a``+``h``,``k``)` becomes `(``1``+``1``,``3``)` or `(2,3)`Directrix:`x=-``a``+``h` becomes `x=-``1``+``1` or `x=0`Start graphing the parabola by plotting the vertex, focus and directrix
Finally, draw a parabola concave right