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3 years ago

How would you graph 4x^2 + 8x -5

Mathematics

1 answer:

professor190 [17]3 years ago

6 0

Answer:

Step-by-step explanation:

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The function f(x)=/-x is shown on the graph

Katyanochek1 [597]

we know that

For the function shown on the graph

The domain is the interval--------> (-∞,0]

$x\leq0$

All real numbers less than or equal to zero

The range is the interval--------> [0,∞)

$y\geq 0$

All real numbers greater than or equal to zero

Statements

case A) The range of the graph is all real numbers less than or equal to $0$

The statement is False

Because the range is all numbers greater than or equal to zero

case B) The domain of the graph is all real numbers less than or equal to $0$

The statement is True

See the procedure

case C) The domain and range of the graph are the same

The statement is False

Because the domain is all real numbers less than or equal to zero and the range is is all numbers greater than or equal to zero

case D) The range of the graph is all real numbers

The statement is False

Because the range is all numbers greater than or equal to zero

therefore

the answer is

The domain of the graph is all real numbers less than or equal to $0$

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4 years ago

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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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