For the given quadratic the vertex is (-4, 6) and the focus (-4, 6.02)
How to find the vertex of the quadratic equation?
Here we have the quadratic equation:
x^2 + 8x - 12y + 88 = 0
Isolating y, we get:
x^2 + 8x + 88 = 12y
(x^2 + 8x + 88)/12 = y
(1/12)*x^2 + (8/12)*x + (88/12) = y
Now, using the formula for the x-value of the vertex we get:
x = -(8/12)/(2*1/12) = -4
To get the y-value of the vertex we evaluate in x = -4, we will get:
y = ((-4)^2 + 8*-4 + 88)/12 = 6
So the vertex is (-4, 6)
Now we want to find the focus, the x-value is the same one as the vertex, and the y-value is:
y = k + 1/4a
Here we have k = y-value o the vertex = 6
a = leading coefficient = 1/12
Then:
y = 6 + 1/4*12 = 6.02
Then the focus is (-4, 6.02)
If you want to learn more about quadratic equations:
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