Which polynomial could have the following graph?

Mathematics

2 answers:

zubka84 [21]3 years ago

8 0

Hi,
Roots are 2, -1,-3.So y=(x-3)(x+1)(x+3)
Answer B

babunello [35]3 years ago

8 0

Answer: $y=(x-2)(x+1)(x+3)$

Step-by-step explanation:

From the given graph, it can be seen that the x intercepts of the graph are

$x=-3,\ x=-1,\ x=2$

It implies the value of y at the above points must be zero.

[x-intercept is of the form (a,0), where the value of y is zero.]

It implies $x=-3,\ x=-1,\ x=2$ are the zeroes of the polynomial graphed.

It implies $(x-2),\ (x+1),\ (x+3)$ are the factors of the polynomial graphed.

[By Factor theorem]

Thus the polynomials should be : $y=(x-2)(x+1)(x+3)$ which is the correct polynomial for the given graph.

You might be interested in

Let T be the plane-2x-2y+z =-13. Find the shortest distance d from the point Po=(-5,-5,-3) to T, and the point Q in T that is cl

GaryK [48]

Answer:

d=10u

Q(5/3,5/3,-19/3)

Step-by-step explanation:

The shortest distance between the plane and Po is also the distance between Po and Q. To find that distance and the point Q you need the perpendicular line x to the plane that intersects Po, this line will have the direction of the normal of the plane $n=(-2,-2,1)$ , then r will have the next parametric equations:

$x=-5-2\lambda\\y=-5-2\lambda\\z=-3+\lambda$

To find Q, the intersection between r and the plane T, substitute the parametric equations of r in T

$-2x-2y+z =-13\\-2(-5-2\lambda)-2(-5-2\lambda)+(-3+\lambda) =-13\\10+4\lambda+10+4\lambda-3+\lambda=-13\\9\lambda+17=-13\\9\lambda=-13-17\\\lambda=-30/9=-10/3$

Substitute the value of $\lambda$ in the parametric equations:

$x=-5-2(-10/3)=-5+20/3=5/3\\y=-5-2(-10/3)=5/3\\z=-3+(-10/3)=-19/3\\$

Those values are the coordinates of Q

Q(5/3,5/3,-19/3)

The distance from Po to the plane

$d=\left| {\to} \atop {PoQ}} \right|=\sqrt{(\frac{5}{3}-(-5))^2+(\frac{5}{3}-(-5))^2+(\frac{-19}{3}-(-3))^2} \\d=\sqrt{(\frac{5}{3}+5))^2+(\frac{5}{3}+5)^2+(\frac{-19}{3}+3)^2} \\d=\sqrt{(\frac{20}{3})^2+(\frac{20}{3})^2+(\frac{-10}{3})^2}\\d=\sqrt{\frac{400}{9}+\frac{400}{9}+\frac{100}{9}}\\d=\sqrt{\frac{900}{9}}=\sqrt{100}\\d=10u$