SOLUTION: y = 5(x+2)^2 + 7. find the focus of the parabola ....

Question 391745: y = 5(x+2)^2 + 7.
find the focus of the parabola ....
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
For this problem use the parabola form, x^2 =4py
y = 5(x+2)^2 + 7
5(x+2)^2=y-7
(x+2)^2 =1/5(y-7)
This is a parabola that opens upwards with the vertex at (-2,7)
line of symmetry = -2
4p =1/5
p=1/20
focus is on the line of symmetry 1/20 above the vertex=7+1/20=141/20
ans: coordinates of focus (-2,141/20)
see graph below

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