By inspection, the parabola is concave down
Therefore general equation is (x-h)^2 = -4a(y-k)
Given that centre is (-4,2), h=-4, k=2
(x+4)^2 = -4a(y-2)
When x=0, y=-16
16 = -4a(-18)
16 = 72a
a = 2/9
Therefore the equation is (x+4)^2 = -8/9(y-2)
Would anyone be able to help me with this question where I have to find the equation of a parabola in vertex form. The information given is that the vertex is (-4,2) and the y intercept is -16. Thank you
By inspection, the parabola is concave down
Therefore general equation is (x-h)^2 = -4a(y-k)
Given that centre is (-4,2), h=-4, k=2
(x+4)^2 = -4a(y-2)
When x=0, y=-16
16 = -4a(-18)
16 = 72a
a = 2/9
Therefore the equation is (x+4)^2 = -8/9(y-2)
HSC 2017: English Adv - Maths Ext 1 - Maths Ext 2 - Chemistry - Physics
ATAR Aim: 97.30
y = a(x-h)^2 + k is the vertex form for a Parabola (note that h,k are the coordinates for the vertex).
Therefore sub (-4,2) in the equation giving you: y = a(x+4)^2 + 2
To get 'a', you must sub a mentioned point in the equation (which in this case is the y-intercept).
After subbing (0,-16 - which is the y-intercept), you should get:
-16 = a(0+4)^2+2
-16 = 16a +2
-18 = 16a
-9/8 = a
Now that you have 'a', you can write the equation as: y = -9/8(x+4)^2 +2
Last edited by _Anonymous; 14 Oct 2017 at 9:26 PM.
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