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1 year ago

What is a cubic polynomial function in standard form with zeros 1 -2 and 2?

Mathematics

1 answer:

Zolol [24]1 year ago

7 0

Answer:

f(x) = x³ - x² - 4x + 4

Step-by-step explanation:

Given the zeros of a polynomial say x = a, x = b, x = c then

(x - a), (x - b), (x - c) are the factors of the polynomial and

f(x) is the product of the factors

here x = 1, x = - 2, x = 2, hence

(x - 1),(x + 2), (x - 2) are the factors and

f(x) = a(x - 1)(x + 2)(x - 2) ← a is a multiplier

let a = 1 and expand the factors

f(x) = (x - 1)(x² - 4)

= x³ - 4x - x² + 4

= x³ - x² - 4x + 4 ← in standard form

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Step-by-step explanation:

5 ÷ 2.5 = 2

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Triangle ABC was dilated with the origin as the center of dilation to create triangle A'B'C'. The triangle was dilated using a s

victus00 [196]

Answer:

$A'(-0.25,0.75)\\\\B'(1.75,-0.25)\\\\C'(-1,-0.25)$

Step-by-step explanation:

A Dilation is defined as a transformation in which the image and the pre-image have the same shape, but their sizes are different.

When the scale factor is greater than 1, the image obtained after the dilation is greater than the pre-image and it is an "Enlargement".

When the scale factor is is between 0 and 1, the image obtained after the dilation is smaller than the pre-image and it is an "Reduction".

In this case you know that the triangle ABC was dilated by this scale factor:

$scale\ factor=\frac{1}{4}$

With the origin as the center of dilation.

Since:

$0$

It is a Reduction.

You can identify that the vertices of the triangle ABC are:

$A(-1,3)\\\\B(7,-1)\\\\C(-4,-1)$

So you need to multiply the coordinates of each vertex of the triangle ABC by $\frac{1}{4}$ , in order to get the coordinates of the triangle A'B'C. Then, you get:

$A'=(\frac{1}{4}(-1),\frac{1}{4}(3))=(-0.25,0.75)\\\\B'(\frac{1}{4}(7),\frac{1}{4}(-1))=(1.75,-0.25)\\\\C'(\frac{1}{4}(-4),\frac{1}{4}(-1))=(-1,-0.25)$