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neonofarm [45]
3 years ago
14

What is the equation of the quadratic graph with a focus of (3, 1) and a directrix of y = 5?

Mathematics
1 answer:
lbvjy [14]3 years ago
4 0

Answer:

The equation of the quadratic graph is f(x)= - (1/8) (x-3)^2 + 3 (second option)

Step-by-step explanation:

Focus: F=(3,1)=(xf, yf)→xf=3, yf=1

Directrix: y=5 (horizontal line), then the axis of the parabola is vertical, and the equation has the form:

f(x)=[1 / (4p)] (x-h)^2+k

where Vertex: V=(h,k)

The directix y=5 must intercept the axis of the parabola at the point (3,5), and the vertex is the midpoint between this point and the focus:

Vertex is the midpoint between (3,5) and (3,1):

h=(3+3)/2→h=6/2→h=3

k=(5+1)/2→k=6/2→k=3

Vertex: V=(h,k)→V=(3,3)

p=yf-k→p=1-3→p=-2

Replacing the values in the equation:

f(x)= [ 1 / (4(-2)) ] (x-3)^2 + 3

f(x)=[ 1 / (-8) ] (x-3)^2 + 3

f(x)= - (1/8) (x-3)^2 + 3

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ANSWER

E. 10 and 640


EXPLANATION


The given series is

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To find the first term of the series we substitute i=1 to get,


a_1 = 5 \times  {2}^{2 \times 1 - 1}


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