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3 years ago

The graph of the function f(x) = −3x2 − 3x + 6 is shown. Which statements describe the graph? Check all that apply.

2 answers:

6 0

Since the coefficient of x^2 is negative, which means that the parabola is facing down. Hence, the vertex is maximum value.
f(-1) = -3(-1)^2 - 3(-1) + 6 = -3(1) - (-3) + 6 = -3 + 3 + 6 = 6
f(-0.5) = -3(-0.5)^2 - 3(-0.5) + 6 = -3(0.25) - (-1.5) + 6 = -0.75 + 1.5 + 6 = 6.75
f(0) = -3(0)^2 - 3(0) + 6 = 0 + 0 + 6 = 6
From the above, the graph stopped increasing at x = -0.5 and started decreasing. Hence, the axis of symmetry is x = -1/2
The domain of all quadatic expression is all real numbers because there is no value of x for which f(x) does not exist.
The range is NOT all real numbers because for instance, there is no value of x for which f(x) = 7.
f(-2) = -3(-2)^2 - 3(-2) + 6 = -3(4) - (-6) + 6 = -12 + 6 + 6 = 0
f(1) = -3(1)^2 - 3(1) + 6 = -3(1) - 3 + 6 = -3 + 3 = 0
Hence, the x-intecepts are at (-2, 0) and (1, 0)
The function is INCREASING from (−∞, 6.75)

Verizon [17]3 years ago

6 0

Answer:

I took the test on edge

Step-by-step explanation:

The vertex is the maximum value.

The axis of symmetry is x = negative one-half.

The domain is all real numbers.

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If <img src="https://tex.z-dn.net/?f=x%20%3D%209%20-%204%5Csqrt%7B5%7D" id="TexFormula1" title="x = 9 - 4\sqrt{5}" alt="x = 9 -

Komok [63]

Observe that

$\left(\sqrt x - \dfrac1{\sqrt x}\right)^2 = \left(\sqrt x\right)^2 - 2\sqrt x\dfrac1{\sqrt x} + \left(\dfrac1{\sqrt x}\right)^2 = x - 2 + \dfrac1x$

Now,

$x = 9 - 4\sqrt5 \implies \dfrac1x = \dfrac1{9-4\sqrt5} = \dfrac{9 + 4\sqrt5}{9^2 - \left(4\sqrt5\right)^2} = 9 + 4\sqrt5$

so that

$\left(\sqrt x - \dfrac1{\sqrt x}\right)^2 = (9 - 4\sqrt5) - 2 + (9 + 4\sqrt5) = 16$

$\implies \sqrt x - \dfrac1{\sqrt x} = \pm\sqrt{16} = \pm 4$

To decide which is the correct value, we need to examine the sign of $\sqrt x - \frac1{\sqrt x}$ . It evaluates to 0 if

$\sqrt x = \dfrac1{\sqrt x} \implies x = 1$

We have

$9 - 4\sqrt5 = \sqrt{81} - \sqrt{16\cdot5} = \sqrt{81} - \sqrt{80} > 0$

Also,

$\sqrt{81} - \sqrt{64} = 9 - 8 = 1$

and $\sqrt x$ increases as $x$ increases, which means

$0 < 9 - 4\sqrt5 < 1$

Therefore for all $0 < x < 1$ ,

$\sqrt x - \dfrac1{\sqrt x} < 0$

For example, when $x=\frac14$ , we get

$\sqrt{\dfrac14} - \dfrac1{\sqrt{\frac14}} = \dfrac1{\sqrt4} - \sqrt4 = \dfrac12 - 2 = -\dfrac32 < 0$

Then the target expression has a negative sign at the given value of $x$ :

$x = 9-4\sqrt5 \implies \sqrt x - \dfrac1{\sqrt x} = \boxed{-4}$

Alternatively, we can try simplifying $\sqrt x$ by denesting the radical. Let $a,b,c$ be non-zero integers ( $c>0$ ) such that

$\sqrt{9 - 4\sqrt5} = a + b\sqrt c$

Note that the left side must be positive.

Taking squares on both sides gives

$9 - 4\sqrt5 = a^2 + 2ab\sqrt c + b^2c$

Let $c=5$ and $ab=-2$ . Then

$a^2+5b^2=9 \implies a^2 + 5\left(-\dfrac2a\right)^2 = 9 \\\\ \implies a^2 + \dfrac{20}{a^2} = 9 \\\\ \implies a^4 + 20 = 9a^2 \\\\ \implies a^4 - 9a^2 + 20 = 0 \\\\ \implies (a^2 - 4) (a^2 - 5) = 0 \\\\ \implies a^2 = 4 \text{ or } a^2 = 5$

$a^2 = 4 \implies 5b^2 = 5 \implies b^2 = 1$

$a^2 = 5 \implies 5b^2 = 4 \implies b^2 = \dfrac45$

Only the first case leads to integer coefficients. Since $ab=-2$ , one of $a$ or $b$ must be negative. We have

$a^2 = 4 \implies a = 2 \text{ or } a = -2$

Now if $a=2$ , then $b=-1$ , and

$\sqrt{9 - 4\sqrt5} = 2 - \sqrt5$

However, $\sqrt5 > \sqrt4 = 2$ , so $2-\sqrt5$ is negative, so we don't want this.

Instead, if $a=-2$ , then $b=1$ , and thus

$\sqrt{9 - 4\sqrt5} = -2 + \sqrt5$

Then our target expression evaluates to

$\sqrt x - \dfrac1{\sqrt x} = -2 + \sqrt5 - \dfrac1{-2 + \sqrt5} \\\\ ~~~~~~~~~~~~ = -2 + \sqrt5 - \dfrac{-2 - \sqrt5}{(-2)^2 - \left(\sqrt5\right)^2} \\\\ ~~~~~~~~~~~~ = -2 + \sqrt5 + \dfrac{2 + \sqrt5}{4 - 5} \\\\ ~~~~~~~~~~~~ = -2 + \sqrt5 - (2 + \sqrt5) = \boxed{-4}$

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2 years ago

Household incomes are normally distributed with a mean of $47,500 and a standard deviation of $12,500. What is the probability o

Romashka [77]

Answer:

Option C: 15.87 is the correct answer

Step-by-step explanation:

Given

Mean = $47500

SD = $12500

We have to find the probability of having $60000 or more

So,

x = $60000

The z-score will be:

$z-score = z = \frac{x-mean}{SD}\\z = \frac{60000-47500}{12500}\\= \frac{12500}{12500}\\= 1$

Looking in the z-score table, we get

P(Z<60000) = 0.8413

In order to find the probability P(Z>60000) => 1-P(Z<60000)

= 1-0.8413

=0.1587

Converting into percentage: 0.1587*100 = 15.87%

Hence,

Option C: 15.87 is the correct answer

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dsp73

Step-by-step explanation:

(-1)⁴ - 11 = 1 - 11 = -10

0⁴ -11 = -11

1⁴ - 11 = -10

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