All categories

4 years ago

If f(x-3)=x^2-4x+5, what's f(1-x)

Mathematics

1 answer:

olganol [36]4 years ago

4 0

Answer:

$\large\boxed{f(1-x)=x^2-4x+5}$

Step-by-step explanation:

$f(x-3)=x^2-4x+5\\\\f((x-3)+3)\to\text{put x + 3 to the equation of the function:}\\\\f(x-3+3)=(x+3)^2-4(x+3)+5\\\text{use the distributive property and}\ (a+b)^2=a^2+2ab+b^2\\f(x)=x^2+2(x)(3)+3^2+(-4)(x)+(-4)(3)+5\\f(x)=x^2+6x+9-4x-12+5\qquad\text{combine like terms}\\f(x)=x^2+(6x-4x)+(9-12+5)\\f(x)=x^2+2x+2\\\\f(1-x)\to\text{put 1 - x to the equation of the funtion f(x):}\\\\f(1-x)=(1-x)^2+2(1-x)+2\\\text{use the distributive property and}\ (a-b)^2=a^2-2ab+b^2$

$f(1-x)=1^2-2(1)(x)+x^2+(2)(1)+(2)(-x)+2\\f(1-x)=1-2x+x^2+2-2x+2\qquad\text{combine like terms}\\f(1-x)=x^2+(-2x-2x)+(1+2+2)\\f(1-x)=x^2-4x+5$

You might be interested in

Multiply 2x(x2 + 5). 2x3 + 10x2 2x3 + 10x 2x2 + 10x 2x3 + 5x

Alex_Xolod [135]

Hope this will help u....

6 0

4 years ago

bryan has some 3 cent stamps and some 4 cent stamps. what is the least number of stamps he can combine so the value of the stamp

SashulF [63]

Answer:

The least number of stamps required is $9$

Step-by-step explanation:

Let the number of $3$ cent stamps be $x$ and $4$ cent stamps be $y$

We have

$3x+4y=33$

The minimum number is obtained when more $4$ cent stamps are used

Here $y$ cannot be greater than $8$ since $\frac{33}{4}$

Substitute $y=8$

$3x+4\times 8=33\\\\3x=1\\\\x=\frac{1}{3}$

Not possible since $x$ is not a fraction

Substitute $y=7$

$3x+4\times 7=33\\\\3x=5\\\\x=\frac{5}{3}$

Not possible since $x$ is not a fraction

Substitute $y=6$

$3x+4\times 6=33\\\\3x=9\\\\x=\frac{9}{3}\\\\=3$

Possible

Hence minimum number of stamps is

$=x+y\\\\=3+6\\\\=9$

5 0

3 years ago

Please show your work this question what made you come to the conclusion. Thank you

AveGali [126]

Answer:

B the range, the x- and y-intercept

Step-by-step explanation:

the domain stays the same : all values of x are possible out of the interval (-infinity, +infinity).

but the range changes, as for the original function y could only have positive values - even for negative x.

the new function has a first term (with b) that can get very small for negative x, and then a subtraction of 2 makes the result negative.

the y-intercept (x=0) of the original function is simply y=1, as b⁰=1.

the y-intercept of the new function is definitely different, because the first term 3×(b¹) is larger than 3, because b is larger than 1. and a subtraction of 2 leads to a result larger than 1, which is different to 1.

the original function has no x-intercept (y=0), as this would happen only for x = -infinity. and that is not a valid value.

the new function has an x-intercept, because the y-values (range) go from negative to positive numbers. any continuous function like this must therefore have an x-intercept (again, y = the function result = 0)

$3 {b}^{x + 1} = 2$