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4 years ago

Use the graph of f(x) = x2 to write an equation for the function represented by each graph.

Mathematics

1 answer:

MariettaO [177]4 years ago

7 0

Answer:

y = -(x + 2)² + 1

Step-by-step explanation:

Parent function given in the graph is a quadratic function,

f(x) = x²

Since, graph is opening downwards transformed function of the preimage will be,

g(x) = -x²

This transformed function is shifted further by 2 units left horizontally and 1 unit upwards.

Therefore, rule for the transformation will be,

g(x) → h[(x + 2), (y + 1)]

By this rule transformed function will be,

h(x) = -(x + 2)²+ 1

Equation of the curve will be,

y = -(x + 2)² + 1

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(12) — (23 + 13i) = Chc

rewona [7]

Answer:

-11 - 13i.

Step-by-step explanation:

(12) — (23 + 13i)

= 12 - 23 - 13i

= -11 - 13i.

4 0

3 years ago

A city planner designs a park that is a quadrilateral with vertices at J(-3,1), K(1,3), L(5,-1), and M(-1,-3). There is an entra

Ksivusya [100]

The Quadrilateral is JKLM,

let $M_{JK}, M_{KL}, M_{LM}, M_{JM},$ be the midpoints of JK, KL, LM and JM respectively.

---------------------------------------------------------------------------------------------------------

Given any 2 point P(m,n) and Q(k,l),

the coordinates of the midpoint of the line segment PQ are given by the formula:

$M_{PQ}=( \frac{m+k}{2} ,
\frac{n+l}{2})$ ,

-------------------------------------------------------------------------------------------------

thus the coordinates of points $M_{JK}, M_{KL}, M_{LM}, M_{JM},$

are as follows:

$M_{JK}= (\frac{-3+1}{2}, \frac{1+3}{2})=(-1,2), \\\\M_{KL}= (\frac{1+5}{2}, \frac{3-1}{2}=(3,1), \\\\M_{LM}= (\frac{5-1}{2}, \frac{-1-3}{2})=(2, -2),\\\\ M_{JM}= (\frac{-3-1}{2}, \frac{1-3}{2})=(-2,-1)$

------------------------------------------------------------------------------------------------

The distance between any 2 points P(a,b) and Q(c,d) in the coordinate plane, is given by the formula:

$|PQ|= \sqrt{ (a-c)^{2} + (b-d)^{2}
}$

------------------------------------------------------------------------------------------------

thus the distances connecting the opposite entrances can be calculated as follows:

$|M_{JK},M_{LM}|= \sqrt{ (-1-2)^{2} + (2-(-2))^{2} }= \sqrt{9+16}=5$

$|M_{KL}M_{JM}|= \sqrt{ (3-(-2))^{2} + (1-(-1))^{2}}= \sqrt{25+4}= \sqrt{29}=5.39$

Thus the total distance of the paths joining the opposite entrances is

5+5.39 units = 50 m + 53.9 m = 104 m (rounded to the nearest meter)

Answer: 104 m

8 0

3 years ago

Can someone please help me with this one

kakasveta [241]

Answer:

mhm

Step-by-step explanation:

4 0

3 years ago

Which equations represent the asymptotes of the hyperbola?

makvit [3.9K]

Answer:

see below

Step-by-step explanation:

The equation of the hyperbola can be written as ...

((x -h)/a)² -((y -k)/b)² = 1

This has asymptotes ...

(x -h)/a ± (y -k)/b = 0

Solving for y, we have ...

y = ±(b/a)(x -h) +k

Filling in the given values a=6, b=8, h=1, k=2, we have ...

y = ±8/6(x -1) +2

$y=\dfrac{\pm4x\mp4+6}{3}\\\\\boxed{y=\dfrac{4x+2}{3}\ \text{and }y=\dfrac{10-4x}{3}}$

4 0

4 years ago

How would the expression x^2+27 be rewritten using sum of cubes?

nydimaria [60]

Assuming the square is not a typo, one can write

$x^2+27=(x^{2/3})^3+3^3=(x^{2/3}+3)\bigg((x^{2/3})^2-3x^{2/3}+9\bigg)$

$=(x^{2/3}+3)\bigg(x^{4/3}-3x^{2/3}+9\bigg)$

6 0

3 years ago