Which is the graph of the function f(x)=x^3+x^2+x+1

Mathematics

1 answer:

34kurt3 years ago

3 0

Answer:

Step-by-step explanation:

Next time, please share the possible answers (which here are graphs).

f(x)=x^3+x^2+x+1 builds upon the cubing function x^3, which starts in Quadrant III and continues into Quadrant I. f(x)=x^3+x^2+x+1 displays the same behavior. Note that if x = 0, f(0) = 1, so the y-intercept is (0, 1).

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Which of the following equations represents the perpendicular bisector of WX graphed below?

kotykmax [81]

The coordinates of the 2 given points are W(-5, 2), and X(5, -4).

First, we find the midpoint M using the midpoint formula:

$\displaystyle{ M_{WX}= (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} )= (\frac{-5+5}{2}, \frac{2+(-4)}{2} )=(0, -1).$

Nex, we find the slope of the line containing M, perpendicular to WX. We know that if m and n are the slopes of 2 parallel lines, then mn=-1.

The slope of WX is $\displaystyle{ m= \frac{y_2-y_1}{x_2-x_1}= \frac{2-(-4)}{-5-5}= \frac{6}{-10}= -\frac{3}{5}$ .

Thus, the slope n of the perpendicular line is $\displaystyle{ \frac{5}{3}$ .

The equation of the line with slope $\displaystyle{ n= \frac{5}{3}$ containing the point M(0, -1) is given by:

$\displaystyle{ y-(-1)=\frac{5}{3}(x-0)$

$\displaystyle{ y+1= \frac{5}{3}x$

$\displaystyle{ 3y+3=5x$

$\displaystyle{ 5x-3y-3=0$

Answer: 5x-3y-3=0

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3 years ago

Can anyone help me out with this ?

Colt1911 [192]

A² + b² = c²

3² + 6² = x²

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Answer: x = 3√5

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3 years ago

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