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# HELP ME ASAP!

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1. There exist constants a, h, and k such that \(3x^2+12x+4=a(x-h)^2+k\)
for all real numbers x. Enter the ordered triple (a,h,k).\

2. Geometrically speaking, a parabola is defined as the set of points that are the same distance from a given point and a given line. The point is called the focus of the parabola and the line is called the directrix of the parabola. Suppose \(P\) is a parabola with focus \((4,3)\) and directrix \(y=1\). The point \((8,6)\) is on \(P\) because \((8,6)\) is \(5\) units away from both the focus and the directrix. If we write the equation whose graph is \(P\) in the form \(y=ax^2+bx+c\), then what is \(a+b+c\)?

Jul 3, 2020

#1
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1.

3x^2 + 12x + 4

=  3(x^2 + 4x) + 4

=  3(x^2 + 4x + 4) + 4 - 12

=  3(x + 2)^2 - 12

a = 3     h = -2     k = -12

2.

The vertex of the parabola will be found half-ways between the focus and the directrix.

The vertex will be the point (4, 1.5).

A formula for the parabola is:  y - k  =  a(x - h)2  where (h, k) are the coordinates of the vertex.

For this situation:  y - 1.5  =  a(x - 4)2.

Since the point  (8, 6)  is on the parabola:     6 - 1.5  =  a(8 - 4)2

--->      4.5  =  a(4)2     --->     4.5  =  16·a     --->     a  =  0.28125

The equation is:  y - 1.5  =  0.28125(x - 4)2.

Multiplying out:   y - 1.5  =  0.28125(x2 - 8x + 16)

y - 1.5  =  0.28125x2 - 2.25x + 4.5

y  =  0.28125x2 - 2.25x + 6

The sum of the coefficients is then 0.28125 - 2.25 + 6 = 129/32

Jul 5, 2020
#2
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Sorry, but both of those answers are wrong.

Jul 5, 2020