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Fofino [41]
3 years ago
7

Given the points (-3,4), (-1,5), (5,10) and (7,11) of a parallelogram. What transformation would translate the parallelogram so

the lower left vertex would be at the origin and all other points would fall in the first quadrant?
A)Translate the parallelogram right three and down four.
B)Translate the parallelogram right one and down five.
C)Translate the parallelogram left five and down ten.
D)Translate the parallelogram left seven and down eleven.
Mathematics
1 answer:
ahrayia [7]3 years ago
5 0

in the given parallelogram the lower left vertex is (-3,4)

to translate the parallelogram in such a way

that its lower left vertex lies at origin means

we have to translate (-3,4) to (0,0)

i.e. (-3,4)+(3,-4)=(0,0)

So (3,-4) is the answer which is represented by option A

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Answer:

\displaystyle f(x) = 3x^2 + 2x + 5\text{ and } g(x) =2x^2 - 4x -2\text{ or } \\ \\  f(x) = 3x^2 + 5 \text{ and } g(x) = x^2 - 4x -2

Step-by-step explanation:

We are given the two functions:

\displaystyle f(x) = 3x^2 + mx +5 \text{ and } g(x) = nx^2 - 4x -2

And that:

\displaystyle h(x) = f(x)\cdot g(x)

With the given conditions that (1, -40) and (-1, 24) satisfy the new function, we want to determine functions <em>f</em> and <em>g</em>.

First, find <em>h: </em>

<em />

<em />\displaystyle \begin{aligned} h(x) & = f(x)\cdot g(x) \\ \\  & = (3x^2 + mx +5)(nx^2 - 4x -2) \end{aligned}

Because (1, -40) and (-1, 24) are points on the graph of <em>h</em>, we have that h(-1) = 40 and h(-1) = 24. In other words:


\displaystyle \begin{aligned} h(1) = -40 & = (3(1)^2 + m(1) +5)(n(1)^2 - 4(1) -2) \\ \\ & = (3 + m +5)(n-4 -2) \\ \\ & = (m+8)(n-6) \\ \\  -40 &= mn-6m+8n-48  \end{aligned}

And:

\displaystyle \begin{aligned} h(-1) = 24 & = (3(-1)^2 + m(-1) +5)(n(-1)^2 -4(-1) -2) \\ \\ & = (3 - m +5)(n + 4 -2) \\ \\ & = (-m+8)(n+2) \\ \\ 24  & = -mn -2m + 8n +16 \end{aligned}

Solve the system of equations. Adding the two equations together yield:

\displaystyle -16 = -8m+16n - 32

Solve for either <em>m</em> or <em>n: </em>

<em />

<em />\displaystyle \begin{aligned} -16 & = -8m + 16n - 32 \\ \\ 16 & = -8m + 16n \\ \\ 8m & = 16n - 16 \\ \\ m & = 2n -2\end{aligned}

Substitute this into one of the two equations above and solve:


\displaystyle \begin{aligned} -40 & = mn - 6m + 8n - 48 \\ \\ 0 & = (2n-2)n -6 (2n-2) + 8n -8 \\ \\ &= (2n^2 - 2n) + (-12n + 12) +8 n - 8 \\ \\ & = 2n^2 -6n + 4 \\ \\ & = n^2 - 3n + 2 \\ \\  &= (n-2)(n-1) \\ \\ &  \end{aligned}

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\displaystyle n = 2 \text{ or } n = 1

Solve for <em>m: </em>

<em />

<em />\displaystyle \begin{aligned}m &= 2n-2 & \text{ or } m & = 2n-2 \\ \\ & = 2(2) - 1 &\text{ or }  & =2(1) -2 \\ \\ &= 2 &\text{ or } & = 0 \end{aligned}

Hence, the values of <em>n</em> and <em>m</em> are either: 2 and 2, respectively; or 1 and 0, respectively.

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