First consider the parent function y = x^3 and its graph. The left side starts in Quadrant III and continues up through Quadrant I. A negative sign in front of the x^3 reflects this original graph in the x-axis; the graph now starts in Quadrant II and descends into Quadrant IV.

The end behavior of the given function is the same as that of the graph of y = -x^3;

The graph begins in Quadrant II and descends into Quadrant IV, down.

8 0

3 years ago

What is the equation for

WARRIOR [948]

The equation of a line that is perpendicular to the given line is y = –4x – 16.

Solution:

The equation of a line given is y = 0.25x – 7

Slope of the given line( $m_1$ ) = 0.25

Let $m_2$ be the slope of the perpendicular line.

Passes through the point (–6, 8).

If two lines are perpendicular then the product of the slopes equal to –1.

$\Rightarrow m_1 \cdot m_2=-1$

$\Rightarrow 0.25\cdot m_2=-1$

$\Rightarrow m_2=\frac{-1}{0.25}$

$\Rightarrow m_2=-4$

Point-slope intercept formula:

$y-y_1=m(x-x_1)$

$x_1=-6, y_1=8$ and $m=-4$

Substitute these in the formula, we get

$y-8=-4(x-(-6))$

$y-8=-4(x+6)$

$y-8=-4x-24$

Add 8 on both sides of the equation.

$y-8+8=-4x-24+8$

$y=-4x-16$

Hence the equation of a line that is perpendicular to the given line is

y = –4x – 16

5 0

3 years ago

Prove {0,2,0,2,0,2,...} does not converge

ElenaW [278]

Recall that a sequence $x_n$ is convergent if and only if $x_n$ is also a Cauchy sequence, which means to say that for any $\varepsilon>0$ , we can find a sufficiently large $N$ for which

$|x_m-x_n|$

whenever both $m$ and $n$ exceed $N$ .

But this never happens if we choose $m=n+1$ and $0$ ; under these conditions, we have

$|x_m-x_n|=|x_{n+1}-x_n|=2\not$

Therefore $x_n$ is not a Cauchy sequence and hence does not converge.

7 0

4 years ago