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____ [38]
4 years ago
15

The graph of g(x) = 5x is a transformation of the graph of f(x) = x. Which of the following describes the transformation?

Mathematics
2 answers:
lianna [129]4 years ago
7 0
<h2>Answer:</h2>

The term which describes the transformation is:

                         Option: d

            d)    stretched vertically    

<h2>Step-by-step explanation:</h2>

We have a parent function i.e. the original function f(x) as:

                 f(x)=x

We know that the transformation of the type:

         f(x) to a f(x)  either vertically stretch or vertically squeeze the function depending on a.

If a>1 then the transformation is a vertical stretch.

and if a<1 then the transformation is a vertical squeeze.

Here a=5>1

Hence, the function g(x) i.e.

g(x)=5x is a vertical stretch of the function f(x).

Alexus [3.1K]4 years ago
6 0

Answer:

The graph is stretched vertically.

Option D is correct.

Step-by-step explanation:

The parent function is f(x) = x

The other function is g(x) = 5x

The value of a describes the vertical stretch of the graph

if a>1 then the graph is vertically stretched

Here g(x) = 5x

a = 5

as a>1, so, the graph is stretched vertically.

Option D is correct.

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Find the value(s) of k such that 4x² + kx + 4 = 0 has 1 rational solution
Blababa [14]

Answer:

If you mean only one rational solution, the answer is

k_1 = 8, k_2 = -8

If you mean at least 1 rational solution, the answer is

k\in (-\infty, -8]\cup[8, \infty)

Step-by-step explanation:

4x^2 + kx + 4 = 0

Let's calculate the discriminant.

\Delta = b^2 - 4ac

\Delta = k^2 -4 \cdot 4 \cdot 4

\Delta = k^2 -64

Now, remember that:

\text{If } \Delta > 0 : \text{2 Real solutions}

\text{If } \Delta = 0 : \text{1 Real solution}

\text{If } \Delta < 0 : \text{No Real solution}

Therefore, I will just consider the first two cases.

k^2 - 64 > 0

and

k^2 -64 = 0

\boxed{\text{For } k^2 - 64 > 0}

k^2 > 64 \Longleftrightarrow k>\pm\sqrt{64}   \Longleftrightarrow k\in (-\infty, -8)\cup(8, \infty)

\boxed{\text{For } k^2 - 64 = 0}

k^2 = 64 \Longleftrightarrow k=\pm\sqrt{64} \implies k_1 = 8, k_2 = -8

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