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3 years ago

Rewrite f(x) = –2(x − 3)2 + 2 from vertex form to standard form. pleae help.

2 answers:

7 0

F(x) = -2(x - 3)² + 2
f(x) = -2(x - 3)(x - 3) + 2
f(x) = -2[x(x - 3) - 3(x - 3)] + 2
f(x) = -2[x(x) - x(3) - 3(x) - 3(-3)] + 2
f(x) = -2(x² - 3x - 3x + 9) + 2
f(x) = -2(x² - 6x + 9) + 2
f(x) = -2(x²) - 2(-6x) - 2(9) + 2
f(x) = -2x² + 12x - 18 + 2
f(x) = -2x² + 12x - 16

zzz [600]3 years ago

6 0

F(x)=-2(x-3)2+2=f(x)=-2(x-3)

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vitfil [10]

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3 years ago

Read 2 more answers

Write the equation -4x^2+9y^2+32x+36y-64=0 in standard form. Please show me each step of the process!

IgorC [24]

Hey there, hope I can help!

$-4x^2+9y^2+32x+36y-64=0$

$\mathrm{Add\:}64\mathrm{\:to\:both\:sides} \ \textgreater \ 9y^2+32x+36y-4x^2=64$

$\mathrm{Factor\:out\:coefficient\:of\:square\:terms} \ \textgreater \ -4\left(x^2-8x\right)+9\left(y^2+4y\right)=64$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}4$
$-\left(x^2-8x\right)+\frac{9}{4}\left(y^2+4y\right)=16$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}9$
$-\frac{1}{9}\left(x^2-8x\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}$

$\mathrm{Convert}\:x\:\mathrm{to\:square\:form}$
$-\frac{1}{9}\left(x^2-8x+16\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)$

$\mathrm{Convert\:to\:square\:form}$
$-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)$

$\mathrm{Convert}\:y\:\mathrm{to\:square\:form}$
$-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y+4\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)$

$\mathrm{Convert\:to\:square\:form}$
$-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)$

$\mathrm{Refine\:}\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right) \ \textgreater \ -\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=1$

$Refine\;once\;more\;-\frac{\left(x-4\right)^2}{9}+\frac{\left(y+2\right)^2}{4}=1$

For me I used
$\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}= 1$
$As\;\mathrm{it\;\:is\:the\:standard\:equation\:for\:an\:up-down\:facing\:hyperbola}$

I know yours is an equation which is why I did not go any further because this is the standard form you are looking for. I would rewrite mine to get my hyperbola standard form. However the one I have provided is the form you need where mine would be.
$\frac{\left(y-\left(-2\right)\right)^2}{2^2}-\frac{\left(x-4\right)^2}{3^2}=1$

Hope this helps!

4 0

4 years ago

Provide an appropriate response. the regression line for the given data is = 1.488x + 60.46. determine the residual of a data po

kap26 [50]

Do you have answer choices ?

7 0

3 years ago

if a rental car agency charges $17 per day plus 10 cents per mile to rent a certain car.Another agency charges $21 per day plus

SashulF [63]

17+(10X)= 21+(5x) is the equation
17 +(10x1)= 21+(5x1)
17+10=21+5
27=27
So your answer is it takes 1 mile per day
Hope its right lol

7 0

4 years ago

Factor. (6x+4) 3(2x+1) 3x + 2 2(3x+2) 2(x+2)

jok3333 [9.3K]

Answer:

2(3x+2) is your answer because you factor out 2. The number 2 is a factor of both 6 and 4 (2*3=6 and 2*2=4).

5 0

3 years ago