Graphing Polynomial Functions

Mathematics

1 answer:

sp2606 [1]3 years ago

8 0

Let's assume

'a' , 'b' are zeros

so, we will find total zeros from graph

We can curve touches x-axis

Let's assume it touches at x=b

so, there would be two zeros at x=b

it also crosses origin

so, another zero is at x=0

it crosses x-axis one more time

Let's assume that zeros at x=a

so, we have got four zeros

x=a,0,b,b

now, we can set up function

$f(x)=(x-0)(x-a)(x-b)(x-b)$

we can also write as

$f(x)=x(x-a)(x-b)^2$ .................Answer

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The line segment AB with endpoints A (-3, 6) and B (9, 12) is dilated with a scale

Gwar [14]

Answer:

C) (-2, 4), (6,8) is the correct answer.

Step-by-step explanation:

Given that line segment AB:

A (-3, 6) and B (9, 12) is dilated with a scale factor 2/3 about the origin.

First of all, let us calculate the distance AB using the distance formula:

$D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Here,

$x_2=9\\x_1=-3\\y_2=12\\y_1=6$

Putting all the values and finding AB:

$AB = \sqrt{(9-(-3))^2+(12-6)^2}\\\Rightarrow AB = \sqrt{(12)^2+(6)^2}\\\Rightarrow AB = \sqrt{144+36}\\\Rightarrow AB = \sqrt{180}\\\Rightarrow AB = 6\sqrt{5}\ units$

It is given that AB is dilated with a scale factor of $\frac{2}{3}$ .

$x_2'=\dfrac{2}{3}\times x_2=\dfrac{2}{3}\times9=6\\x_1'=\dfrac{2}{3}\times x_1=\dfrac{2}{3}\times-3=-2\\y_2'=\dfrac{2}{3}\times y_2=\dfrac{2}{3}\times 12=8\\y_1'=\dfrac{2}{3}\times y_1=\dfrac{2}{3}\times 6=4$

So, the new coordinates are A'(-2,4) and B'(6,8).

Verifying this by calculating the distance A'B':

$A'B' = \sqrt{(6-(-2))^2+(8-4)^2}\\\Rightarrow A'B' = \sqrt{(8)^2+(4)^2}\\\Rightarrow A'B' = \sqrt{64+16}\\\Rightarrow A'B' = \sqrt{80}\\\Rightarrow A'B' = 4\sqrt{5}\ units = \dfrac{2}{3}\times AB$