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Graph Parabolas (Focus and Directrix) 1Graph Parabolas (Focus and Directrix) 1
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Question 1 of 5
1. Question
Which of the following will most likely be the graph of`x^2=4ay`Hint
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In a quadratic equation, if the coefficient of `y` is positive, the parabola will be concave up. If it is negative, the parabola will be concave downNotice that there is no constant term (term without a variable).`x^2` `=` `4ay` This means that the vertex is at `(0,0)`, the originAlso, take note of the following components:Focus `(0,a)`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 `y`.`x^2` `=` `4a``y` Hence, the parabola is concave up.Therefore, it would make sense to say that the graph for `x^2=4ay` is
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Question 2 of 5
2. Question
Sketch the graph of `x^2=16y`Hint
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Standard Form (Concave Up)
`x^2=4``a``y`Focus
`(0,``a``)`Standard Form (Concave Down)
`x^2=-4``a``y`Focus
`(0,-``a``)`First, determine which standard form applies to the given equation.`x^2` `=` `16y` Since the coefficient of `y` is positive, use `x^2=4``a``y`This also means that the parabola will be concave upSolve for the focal length, `a`, by comparing the given equation to the standard form`x^2` `=` `4a``y` `x^2` `=` `16``y` `4a` `=` `16` `4a``divide4` `=` `16``divide4` Divide both sides by `4` `a` `=` `4` Identify the focus and directrix.Focus:`(0,``a``)` becomes `(0,``4``)`Directrix:`y=-``a` becomes `y=-``4`Start graphing the parabola by plotting the focus and directrix
Finally, draw a parabola that is concave up
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Question 3 of 5
3. Question
Sketch the graph of `x^2=-8y`Hint
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Standard Form (Concave Up)
`x^2=4``a``y`Focus
`(0,``a``)`Standard Form (Concave Down)
`x^2=-4``a``y`Focus
`(0,-``a``)`First, determine which standard form applies to the given equation.`x^2` `=` `-8y` Since the coefficient of `y` is negative, use `x^2=-4``a``y`This also means that the parabola will be concave downSolve for the focal length, `a`, by comparing the given equation to the standard form`x^2` `=` `-4a``y` `x^2` `=` `-8``y` `-4a` `=` `-8` `-4a``divide-4` `=` `-8``divide-4` Divide both sides by `-4` `a` `=` `2` Identify the focus and directrix.Focus:`(0,-``a``)` becomes `(0,-``2``)`Directrix:`y=``a` becomes `y=``2`Start graphing the parabola by plotting the focus and directrix<
Finally, draw a parabola that is concave down
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Question 4 of 5
4. Question
Sketch the graph of `y^2=20x`Hint
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Standard Form (Concave Right)
`y^2=4``a``x`Focus
`(``a``,0)`Standard Form (Concave Left)
`y^2=-4``a``x`Focus
`(-``a``,0)`First, determine which standard form applies to the given equation.`y^2` `=` `20x` Since the coefficient of `x` is positive, use `y^2=4``a``x`This also means that the parabola will be concave rightSolve for the focal length, `a`, by comparing the given equation to the standard form`y^2` `=` `4a``x` `y^2` `=` `20``x` `4a` `=` `20` `4a``divide4` `=` `20``divide4` Divide both sides by `4` `a` `=` `5` Identify the focus and directrix.Focus:`(``a``,0)` becomes `(``5``,0)`Directrix:`x=-``a` becomes `x=-``5`Start graphing the parabola by plotting the focus and directrix
Finally, draw a parabola that is concave right
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Question 5 of 5
5. Question
Sketch the graph of `y^2=-2x`Hint
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Standard Form (Concave Right)
`y^2=4``a``x`Focus
`(``a``,0)`Standard Form (Concave Left)
`y^2=-4``a``x`Focus
`(-``a``,0)`First, determine which standard form applies to the given equation.`y^2` `=` `-2x` Since the coefficient of `x` is negative, use `y^2=-4``a``x`This also means that the parabola will be concave leftSolve for the focal length, `a`, by comparing the given equation to the standard form`y^2` `=` `-4a``x` `y^2` `=` `-2``x` `-4a` `=` `-2` `-4a``divide-4` `=` `-2``divide-4` Divide both sides by `-4` `a` `=` `1/2` Identify the focus and directrix.Focus:`(-``a``,0)` becomes `(-``1/2``,0)`Directrix:`x=``a` becomes `x=``1/2`Start graphing the parabola by plotting the focus and directrix
Finally, draw a parabola that is concave left
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