For [tex]f(x) = 4x + 1[/tex] and (x) = [tex]g(x)=

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cocoj3205

23.05.2020

Mathematics

Answer is C
Answer is C
Answer is C

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Ответ:

taylordalton93

09.02.2022

Mathematics

The standard form is (x - h)^2 = 4p (y - k), where the focus is (h, k + p) and the directrix is y = k - p

We know that the focus is (0, -2) which results in h=0 and k+p=-2.
We also know that the directrix of the parabola is represented by y = 2 so k-p=2

k-p=2
k+p=-2

If we add those 2 equations:
2k=0
k=0
p=-2

Know we replace in the standard form:
(x-0)^2 = 4*(-2)*(y-0)
x^2 = -8*y

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For $f(x) = 4x + 1$ and (x) = $g(x)= x^{2} -5,$ find $(\frac{g}{f}) (x)$ a. $\frac{x^{2} - 5 }{4x +1 },x$ ≠ $-\frac{1}{4}$ b. x $\frac{4 x +1 }{x^{2} - 5}, x$ ≠ ± $\sqrt[]{5}$ c. $\frac{4x +1}{x^{2} -5}$ d. $\frac{x^{2} -5 }{4x + 1}$

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For and (x) = find a. ≠ b. x ≠ ± c. d.

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For $f(x) = 4x + 1$ and (x) = $g(x)= x^{2} -5,$ find $(\frac{g}{f}) (x)$ a. $\frac{x^{2} - 5 }{4x +1 },x$ ≠ $-\frac{1}{4}$ b. x $\frac{4 x +1 }{x^{2} - 5}, x$ ≠ ± $\sqrt[]{5}$ c. $\frac{4x +1}{x^{2} -5}$ d. $\frac{x^{2} -5 }{4x + 1}$