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4 years ago

Which transformations take f(x)=3x+1 to g(x)=−3x-5 ?

1 answer:

8 0

reflection over the x-axis and a translation 4 units down Refelcting f(x) over the x axis gives -f(x)=-3x-1 Subtracting a constant from -f(x) moves the graph of -f(x) that many units down. -f(x)-4=-3x-5=g(x) This shows that g(x) is obtained by reflecting f(x) over the x-axis and then translating 4 units down.

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goblinko [34]

Answer:

(0,0) (-2,1) (-4,2) (-6,3) (-8,4)

Step-by-step explanation:

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3 years ago

Read 2 more answers

A parabola can be drawn given a focus of

Volgvan

Answer:

$y=-\frac{1}{4}(x-2)^{2}+9$

Step-by-step explanation:

Any point on a given parabola is equidistant from focus and directrix.

Given:

Focus of the parabola is at $(2,8)$ .

Directrix of the parabola is $y=10$ .

Let $(x,y)$ be any point on the parabola. Then, from the definition of a parabola,

Distance of $(x,y)$ from focus = Distance of $(x,y)$ from directrix.

Therefore,

$\sqrt{(x-2)^{2}+(y-8)^{2}}=|y-10|$

Squaring both sides, we get

$(x-2)^{2}+(y-8)^{2}=(y-10)^{2}\\(x-2)^{2}=(y-10)^{2}-(y-8)^{2}\\(x-2)^{2}=(y-10+y-8)(y-10-(y-8))...............[\because a^{2}-b^{2}=(a+b)(a-b)]\\(x-2)^{2}=(2y-18)(y-10-y+8)\\(x-2)^{2}=2(y-9)(-2)\\(x-2)^{2}=-4(y-9)\\y-9=-\frac{1}{4}(x-2)^{2}\\y=-\frac{1}{4}(x-2)^{2}+9$

Hence, the equation of the parabola is $y=-\frac{1}{4}(x-2)^{2}+9$ .

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