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2 years ago

Consider the function f(x)=x^2+2x-8

Mathematics

1 answer:

sattari [20]2 years ago

8 0

Answer:

The answer to your question is below

Step-by-step explanation:

Data

function f(x) = x² + 2x - 8

-See the graph below

1) x-intercepts

We observe in the graph that there are no x-intercepts.

2) y-intercepts

We observe in the graph that the y-intercept is (0, 8). Point A

3) axis of symmetry

The axis of symmetry is the line that divides the parabola into two equal parts. This line is x = -1 (blue line)

4) The vertex is the lowest point of the parabola, this point is (-1, 7)

Point B in the graph.

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zavuch27 [327]

Answer:

h, j2, f, g, j1, i, k, l (ell)

Step-by-step explanation:

The horizontal asymptote is the constant term of the quotient of the numerator and denominator functions. Generally, it it is the coefficient of the ratio of the highest-degree terms (when they have the same degree). It is zero if the denominator has a higher degree (as for function f(x)).

We note there are two functions named j(x). The one appearing second from the top of the list we'll call j1(x); the one third from the bottom we'll call j2(x).

The horizontal asymptotes are ...

h(x): 16x/(-4x) = -4
j1(x): 2x^2/x^2 = 2
i(x): 3x/x = 3
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j2(x): 3x^2/-x^2 = -3
f(x): 0x^2/(12x^2) = 0
k(x): 5x^2/x^2 = 5

So, the ordering least-to-greatest is ...

h (-4), j2 (-3), f (0), g (1), j1 (2), i (3), k (5), l (7.5)

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2 years ago

Which pairs of variables would most likely have a positive association? Choose all that apply.

alex41 [277]

Answer:

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Step-by-step explanation:

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2 years ago

how much would $150 invested at 8% interest compounded anually be worth after 13 years? A(t)=P(1+r/n)^

PIT_PIT [208]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$150\\
r=rate\to 8\%\to \frac{8}{100}\to &0.08\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &13
\end{cases}
\\\\\\
A=150\left(1+\frac{0.08}{1}\right)^{1\cdot 13}\implies A=150(1.08)^{13}\implies A\approx 407.9435589$

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