All categories

4 years ago

Also these 2 :) determining the vertex and shifts/translations

Mathematics

1 answer:

Mrrafil [7]4 years ago

6 0

Answer:

The answer to your question is below

Step-by-step explanation:

A) f(x) = x² + 4x + 1

Find the vertex

y = x² + 4x + 1

-Subtract 1 in both sides

y - 1 = x² + 4x

-Complete the perfect trinomial square in the right side

y - 1 + (2)² = x² + 4x + (2)²

-Simplify

y - 1 + 4 = (x + 2)²

y + 3 = (x + 2)²

-Find the vertex

V = (-2, -3)

f(x) = x²

g(x) = (x - 4)² - 2 See the picture below

Shifted right 4 units

Shifted down 2 units

You might be interested in

A triangle with vertices (2, 9), (5, 1), and (1, 4) is dilated by a scale factor of 4. Which matrix expression represents the tr

NARA [144]

<h2>Answer:</h2>

The matrix expression that represents the transformation is:

Option: C

$4\cdot \left[\begin{array}{ccc}2&9\\-5&1\\1&-4\end{array}\right]$

<h2>Step-by-step explanation:</h2>

We are given a vertices of a triangle as:

(2, 9), (-5, 1), and (1,- 4)

This means that in the matrix representation these vertices could be represented as:

$\left[\begin{array}{ccc}2&9\\-5&1\\1&-4\end{array}\right]$

Now, when these vertices are dilated by a scale factor of 4, the change in vertices are:

Each of the vertices are multiplied by 4.

i.e. each of the entry in the matrix is multiplied by 4.

or we can take common 4 from the matrix to obtain the transformed vertices in the form as:

$4\cdot \left[\begin{array}{ccc}2&9\\-5&1\\1&-4\end{array}\right]$

4 0

4 years ago

A CIRCLE HAS A CIRCUMFENCE 43.96 YD FIND THE RADUIS OF THE CIRCLE

sleet_krkn [62]

Answer:

≈ 7

Step-by-step explanation:

8 0

3 years ago

Which of the following represents the zeros of f(x) = 2x3 − 5x2 − 28x + 15?

likoan [24]

$\mathrm{Use\:the\:rational\:root\:theorem}$

$a_0=15,\:\quad a_n=2$

$\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:3,\:5,\:15,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1,\:2$

$\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:3,\:5,\:15}{1,\:2}$

$-\frac{3}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+3$

$\mathrm{Compute\:}\frac{2x^3-5x^2-28x+15}{x+3}\mathrm{\:to\:get\:the\:rest\:of\:the\:eqution:\quad }2x^2-11x+5$

$=\left(x+3\right)\left(2x^2-11x+5\right)$

$Factor: 2x^2-11x+5$

$2x^2-11x+5=\left(2x^2-x\right)+\left(-10x+5\right)$

$=x\left(2x-1\right)-5\left(2x-1\right)$

$2x^3-5x^2-28x+15=\left(x+3\right)\left(x-5\right)\left(2x-1\right)$

$\left(x+3\right)\left(x-5\right)\left(2x-1\right)=0$

$\mathrm{Using\:the\:Zero\:Factor\:Principle:}$

thus zeros of f(x) is

$x=-3,\:x=5,\:x=\frac{1}{2}$

5 0

3 years ago

Read 2 more answers

Using the quadratic formula to solve x2 = 5 – x, what are the values of x?

Ivenika [448]

Answer:

x²=5-x

x²+x-5=0

1x²+1x²-5=0

[ax²+bx²+c=0]

∆=b²-4ac

∆=1²-4(-5)(1)=1+20=21

x=(-b±√∆)/2a

x=(-1±√21)/2

meaning:

x=(-1+√21)/2 or x=(-1-√21)/2

4 0

2 years ago

Can anyone help me with calculus??

Rina8888 [55]

Answer:

First one is B Second one is D thats all i know and nice candy cane:)

3 0

3 years ago