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yaroslaw [1]
3 years ago
15

In the following graph of F(x) = -(2/3)(x-3)2 + 2 is the preimage of a transformation of G(x)which is the image. What is the map

ping statement for the function G(x)?

Mathematics
1 answer:
julia-pushkina [17]3 years ago
4 0

Question:

In the following graph of f(x)=-\frac{2}{3} (x-3)^2+2 is the preimage of a transformation of G(x)which is the image. What is the mapping statement for the function G(x)?

The image of the transformed graph is attached below:

Answer:

The function g(x) is g(x)=-\frac{2}{3} (x+1)^2+1

Explanation:

The function is f(x)=-\frac{2}{3} (x-3)^2+2

Let us determine the transformed equation of the function from the graph.

From the graph, we can see that the function g(x) is shifted 1 unit downwards and shifted 4 units to the left.

Thus, the function g(x) can be written as

g(x)=-\frac{2}{3} (x-3+4)^2+2-1

Simplifying, we have,

g(x)=-\frac{2}{3} (x+1)^2+1

Hence, the function g(x) is g(x)=-\frac{2}{3} (x+1)^2+1

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Determine the equation of the line that is perpendicular to the given line, through the given point.
exis [7]

Answer:

The equation of the line is y + 7 = -\frac{1}{3}(x - 9)

Step-by-step explanation:

Equation of a line:

The equation of line, in point-slope form, is given by:

y - y_0 = m(x - x_0)

In which m is the slope and the point is (x_0, y_0)

Perpendicular lines:

If two lines are perpendicular, the multiplication of their slopes is -1.

Through (9,-7)

This means that x_0 = 9, y_0 = -7

So

y - y_0 = m(x - x_0)

y - (-7) = m(x - 9)

y + 7 = m(x - 9)

Perpendicular to y = 3x + 4

This line has slope 3, so:

3m = -1

m = -\frac{1}{3}

Thus

y + 7 = m(x - 9)

y + 7 = -\frac{1}{3}(x - 9)

4 0
3 years ago
What are the solutions of x^2-2x+5=0
sweet [91]

The solution of x^{2}-2 x+5=0 are 1 + 2i and 1 – 2i

<u>Solution:</u>

Given, equation is x^{2}-2 x+5=0

We have to find the roots of the given quadratic equation

Now, let us use the quadratic formula

x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}  --- (1)

<em><u>Let us determine the nature of roots:</u></em>

Here in x^{2}-2 x+5=0 a = 1 ; b = -2 ; c = 5

b^2 - 4ac = 2^2 - 4(1)(5) = 4 - 20 = -16

Since b^2 - 4ac < 0 , the roots obtained will be complex conjugates.

Now plug in values in eqn 1, we get,

x=\frac{-(-2) \pm \sqrt{(-2)^{2}-4 \times 1 \times 5}}{2 \times 1}

On solving we get,

x=\frac{2 \pm \sqrt{4-20}}{2}

x=\frac{2 \pm \sqrt{-16}}{2}

x=\frac{2 \pm \sqrt{16} \times \sqrt{-1}}{2}

we know that square root of -1 is "i" which is a complex number

\begin{array}{l}{\mathrm{x}=\frac{2 \pm 4 i}{2}} \\\\ {\mathrm{x}=1 \pm 2 i}\end{array}

Hence, the roots of the given quadratic equation are 1 + 2i and 1 – 2i

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3 years ago
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