Can anyone help me? Which function is graphed on the right?

Mathematics

2 answers:

Kazeer [188]4 years ago

7 0

<h2>Answer:</h2>

$y=2^{\left(x-2\right)}+3$

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

To solve this problem, we need to start with the parent function of the exponential function, which is $f(x)=a^x$ , where $a$ is the base. In our problem, $a=2$ , so our parent function here is $y=2^x$ . Then, we need to perform some transformations to our parent function. Thus:

1. Vertical shrink:

A vertical shrink is a nonrigid transformation because the graph of the function get a distortion in the shape, so this transformation is as follows:

$g(x)=cf(x)$

where $c$ in this problem equals 0.25 because:

$y=0.25(2^x) \\ \\ y=\frac{1}{4}(2^x) \\ \\ y=\frac{2^x}{2^2} \\ \\ y=2^{(x-2)}$

2. Vertical shift:

The graph of the function $y=2^{(x-2)}$ get a vertical shift given by:

$y=2^{(x-2)}+3$

So the graph is shifted 3 units up. So the result is the graph shown above.

uysha [10]4 years ago

3 0

Answer:

The answer is C

Step-by-step explanation:

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A rocket is fired upward from some initial distance above the ground. Its height in feet, h,above the ground, t seconds after

kirill115 [55]

Check the picture below.

$\bf \textit{ vertex of a vertical parabola, using coefficients}\\\\
\begin{array}{lccclll}
h(t) = &{{ -16}}t^2&{{ +96}}t&{{ +2560}}\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad 
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)$

so.. the maximum point is at $\bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}~~~ft$

and it happened after $\bf -\cfrac{{{ b}}}{2{{ a}}}~~~seconds$

--------------------------------------------------------------------------

now, when the rocket reaches the ground, as you saw on the picture, y = 0, thus

$\bf h(t)=-16t^2+96t+2560\implies 0=-16t^2+96t+2560
\\\\\\
0=-16(t^2-6t-160)\implies 0=t^2-6t-160
\\\\\\
0=(t-16)(t+10)\implies t=
\begin{cases}
\boxed{16}\\
-10
\end{cases}$

"t" is an amount in seconds, for this specific phenomena, it cannot be negative, so -10 is not a feasible value.