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Veseljchak [2.6K]
3 years ago
5

What is the equation of the following graph in vertex form? Courtesy of Texas Instruments A. y = (x − 4)2 − 4 B. y = (x + 4)2 −

4 C. y = (x + 2)2 + 6 D. y = (x + 2)2 + 12

Mathematics
1 answer:
Shtirlitz [24]3 years ago
6 0

Answer:

A: y = (x − 4)^2 − 4

Step-by-step explanation:

vertex=(4.-4)

A: y = (x − 4)^2 − 4

y=x^2-8x+16-4

y=x^2-8x+12 (a=1,b=-8,c=12)

the y intercept is (0,12)

vertex ( h, k)

h=-b/2a ⇒ h=-(-8)/2=4

plug the value of h in the equation y=x^2-8x+12

k=4²-8(4)+12

k=16-32+12

k=-4

v(4,-4)

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Write the recursive formula for -34, -38, -42, -46, -50 ...
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Which statement describes the inverse of m(x) = x^2 – 17x?
DochEvi [55]

Given:

The function is

m(x)=x^2-17x

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The inverse of the given function.

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We have,

m(x)=x^2-17x

Substitute m(x)=y.

y=x^2-17x

Interchange x and y.

x=y^2-17y

Add square of half of coefficient of y , i.e., \left(\dfrac{-17}{2}\right)^2 on both sides,

x+\left(\dfrac{-17}{2}\right)^2=y^2-17y+\left(\dfrac{-17}{2}\right)^2

x+\left(\dfrac{17}{2}\right)^2=y^2-17y+\left(\dfrac{17}{2}\right)^2

x+\left(\dfrac{17}{2}\right)^2=\left(y-\dfrac{17}{2}\right)^2        [\because (a-b)^2=a^2-2ab+b^2]

Taking square root on both sides.

\sqrt{x+\left(\dfrac{17}{2}\right)^2}=y-\dfrac{17}{2}

Add \dfrac{17}{2} on both sides.

\sqrt{x+\left(\dfrac{17}{2}\right)^2}+\dfrac{17}{2}=y

Substitute y=m^{-1}(x).

m^{-1}(x)=\sqrt{x+(\dfrac{189}{4}})+\dfrac{17}{2}

We know that, negative term inside the root is not real number. So,

x+\left(\dfrac{17}{2}\right)^2\geq 0

x\geq -\left(\dfrac{17}{2}\right)^2

Therefore, the restricted domain is x\geq -\left(\dfrac{17}{2}\right)^2 and the inverse function is m^{-1}(x)=\sqrt{x+(\dfrac{189}{4}})+\dfrac{17}{2}.

Hence, option D is correct.

Note: In all the options square of \dfrac{17}{2} is missing in restricted domain.

7 0
3 years ago
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