Answer: the function g(x) has the smallest minimum y-value.

Explanation:

1) The function f(x) = 3x² + 12x + 16 is a parabola.

The vertex of the parabola is the minimum or maximum on the parabola.

If the parabola open down then the vertex is a maximum, and if the parabola open upward the vertex is a minimum.

The sign of the coefficient of the quadratic term tells whether the parabola opens upward or downward.

When such coefficient is positive, the parabola opens upward (so it has a minimum); when the coefficient is negative the parabola opens downward (so it has a maximum).

Here the coefficient is positive (3), which tells that the vertex of the parabola is a miimum.

Then, finding the minimum value of the function is done by finding the vertex.

I will change the form of the function to the vertex form by completing squares:

Given: 3x² + 12x + 16

Group: (3x² + 12x) + 16
Common factor: 3 [x² + 4x ] + 16
Complete squares: 3[ ( x² + 4x + 4) - 4] + 16
Factor the trinomial: 3 [(x + 2)² - 4] + 16
Distributive property: 3 (x + 2)² - 12 + 16
Combine like terms: 3 (x + 2)² + 4

That is the vertex form: A(x - h)² + k, whch means that the vertex is (h,k) = (-2, 4).

Then the minimum value is 4 (when x = - 2).

2) The othe function is <span>g(x)= 2 *sin(x-pi)
</span>

The sine function goes from -1 to + 1, so the minimum value of sin(x - pi) is - 1.

When you multiply by 2, you just increased the amplitude of the function and obtain the new minimum value is 2 (-1) = - 2

Comparing the two minima, you have 4 vs - 2, and so the function g(x) has the smallest minimum y-value.

7 0

3 years ago

Find the remaining trigonometric ratios of θ if csc(θ) = -6 and cos(θ) is positive

VikaD [51]

Now, the cosecant of θ is -6, or namely -6/1.

however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.

we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now

$\bf csc(\theta)=-6\implies csc(\theta)=\cfrac{\stackrel{hypotenuse}{6}}{\stackrel{opposite}{-1}}\impliedby \textit{let's find the \underline{adjacent side}}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\pm\sqrt{6^2-(-1)^2}=a\implies \pm\sqrt{35}=a\implies \stackrel{IV~quadrant}{+\sqrt{35}=a}$

recall that

$\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad\qquad 
cos(\theta)=\cfrac{adjacent}{hypotenuse}
\\\\\\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{adjacent}{opposite}
\\\\\\
% cosecant
csc(\theta)=\cfrac{hypotenuse}{opposite}
\qquad \qquad 
% secant
sec(\theta)=\cfrac{hypotenuse}{adjacent}$

therefore, let's just plug that on the remaining ones,

$\bf sin(\theta)=\cfrac{-1}{6}
\qquad\qquad 
cos(\theta)=\cfrac{\sqrt{35}}{6}
\\\\\\
% tangent
tan(\theta)=\cfrac{-1}{\sqrt{35}}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{\sqrt{35}}{1}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}$

now, let's rationalize the denominator on tangent and secant,

$\bf tan(\theta)=\cfrac{-1}{\sqrt{35}}\implies \cfrac{-1}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{-\sqrt{35}}{(\sqrt{35})^2}\implies -\cfrac{\sqrt{35}}{35}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}\implies \cfrac{6}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{6\sqrt{35}}{(\sqrt{35})^2}\implies \cfrac{6\sqrt{35}}{35}$

3 0

3 years ago

Find the circumference of the pizza to the nearest tenth.

mote1985 [20]

Answer:

C=2πr

r=D/2=10/2=5

C=2×3.14×5

C=31.4in

7 0

3 years ago

A migrating bird flies 539 miles in 11 hours. How many miles does it fly in 3 hours?

joja [24]

It would fly 147 miles in 3 hours

8 0

2 years ago

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