SOLUTION: The equation of a parabola is given.
y=14x2−3x+18
What are the coordinates of the focus of the parabola?
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Question 1081913: The equation of a parabola is given.
y=14x2−3x+18
What are the coordinates of the focus of the parabola?
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
Write in vertex form.
vertex is at -b/2a=3/28 for x and y=(14*9/28^2)x^2-9/28+18=126/784-252/784+18=17.84
ax^2+bx+c=f(x)
f(1)=29
f(-1)=35
y=a(x-0.107)^2+17.84, 14(x-0.107)^2+17.84
29=a(.893)^2+17.84
11.16/(0.893)^2=a=14, which it should be.
focus is at (h, k+1/(4a)); k+1/(4a)=17.84+1/56=17.86
that is at (0.107, 17.86)
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