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Question 427800: i need to know how to identify the coordinates of the vertex and focus, the equations of the axis of symmetry and directrix and the direction of opening of the parabola withe the given equation. and then i need to know how to graph the equation..equation is y=-(x+3)^2+4..can you help me?
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! i need to know how to identify the coordinates of the vertex and focus, the equations of the axis of symmetry and directrix and the direction of opening of the parabola with the given equation. and then i need to know how to graph the equation..equation is y=-(x+3)^2+4..can you help me.
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y=-(x+3)^2+4
This is a parabola of the form, y=-(x-h)^2+k, with (h,k) being the (x,y) coordinates of the vertex.
So the vertex of the given parabola is at (-3,4). Because the coefficient of the (x+3) is negative, the parabola opens downward. If the coefficient were positive, the parabola would open upwards.
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To find the focus and directrix, you need to change the equation into another form, -(x-h)^2=4p(y-k), (h,k) still representing the (x,y) coordinates of the vertex. Note that 4p is the coefficient of the (y-k) term. For the given equation, 4p=1. so p=1/4. On the axis of symmetry, x=-3, the focus is 1/4 unit below the vertex at (-3,3+3/4)and the directrix is a horizontal line above the vertex, that is, y=4+1/4
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To graph the equation, you already have one point, the vertex at (-3,4). All you need is another point and noting that the curve is symmetrical about the axis of symmetry.
You should get something similar to the graph below.
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