Mathematics, 29.07.2019 22:40 jngonzo1226

Using a directrix of y = −3 and a focus of (2, 1), what quadratic function is created? f(x) = one eighth (x − 2)2 − 1 f(x) = −one eighth (x + 2)2 − 1 f(x) = one half (x − 2)2 + 1 f(x) = −one half (x + 2)2 − 1

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Answer from: blondielocks2002

f(x) = one eighth (x - 2)^2 - 1
Since a parabola is the curve such that all points on the curve have the same distance from the directrix as the distance from the point to the focus.With that in mind, we can quickly determine 3 points on the parabola. The 1st point will be midway between the focus and the directrix, So: (2, (1 + -3)/2) = (2, -2/2) = (2,-1). The other 2 points will have the same y-coordinate as the focus, but let offset on the x-axis by the distance from the focus to the directrix. Since the distance is (1 - -3) = 4, that means the other 2 points will be (2 - 4, 1) and (2 + 4, 1) which are (-2, 1) and (6, 1). The closest point to the focus will have the same x-coordinate as the focus, so the term will be (x-2)^2. This eliminates the functions "f(x) = -one eighth (x + 2)^2 - 1" and "f(x) = -one half (x + 2)^2 - 1" from consideration since their x term is incorrect, leaving only "f(x) = one eighth (x - 2)^2 - 1" and "f(x) = one half (x - 2)^2 + 1" as possible choices. Let's plug in the value 6 for x and see what y value we get from squaring (x-2)^2. So: (x-2)^2 (6-2)^2 = 4^2 = 16 Now which option is equal to 1? Is it one eighth of 16 minus 1, or one half of 16 plus 1? 16/8 - 1 = 2 - 1 = 1 16/2 + 1 = 8 + 1 = 9 Therefore the answer is "f(x) = one eighth (x - 2)^2 - 1"

Answer from: travisvb

f(x) = one eigth (x-2)^2 - 1

Step-by-step explanation:

We have. focus = (2,1) and directrix y = -3.

Let (x,y) be any point on the curve.

First, we will find the distance between (x,y) and the focus (2,1).

i.e. distance = $\sqrt{(x-2)^{2} +(y-1)^{2} }$

Now, we will find the distance between (x,y) and directrix y = -3.

i.e. distance = | y - (-3) | = | y + 3 |

To find the quadratic function, we will equate both the distances and simplify the equation.

i.e. $\sqrt{(x-2)^{2} +(y-1)^{2} }$ = | y + 3 |

Squaring both sides, we get,

$(x-2)^{2}+(y-1)^{2}=(y+3)^{2}$

i.e. $x^{2}-4x-8y-4=0$

i.e. $8y=x^{2}-4x-4$

i.e. $y=\frac{x^{2} }{8} -\frac{x}{2} -\frac{1}{2}$

The option f(x) = one eigth (x-2)^2 - 1 will give the y obtained above after simplifying.

Answer from: Quest

Ithink the answer will be d yes

Answer from: Quest

Ican’t sorry i would if i could

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Using a directrix of y = −3 and a focus of (2, 1), what quadratic function is created? f(x) = one e...

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