All categories

3 years ago

Write the equation of a linear function for which f=(-4)=0 and f=(0)=-2

Mathematics

1 answer:

densk [106]3 years ago

6 0

Answer:

y = -0.5x - 2

Step-by-step explanation:

equation: f(x) = mx + b

f(-4) : -4m + b = 0 ...(1)

f(0): 0m + b = -2 b = -2 ... plug in (1)

-4m - 2 = 0

m = - 1/2 = - 0.5

linear equation: y = -0.5x - 2

You might be interested in

Pleas help me sovle this

frozen [14]

What exactly are we solving?? the rest of the angles or the perimeter?

4 0

2 years ago

If c (x) = StartFraction 5 Over x minus 2 EndFraction and d(x) = x + 3, what is the domain of (cd)(x)?

Volgvan

Answer:

The domain of the function (cd)(x) will be all real values of x except x = 2.

Step-by-step explanation:

The two functions are $c(x) = \frac{5}{x - 2}$ and d(x) = x + 3

So, (cd)(x) = $(\frac{5}{x - 2})(x + 3) = \frac{5(x + 3)}{x - 2}$

Then, for x = 2 the function (cd)(x) will be undefined as zero in the denominator will make the function (cd)(x) undefined.

Therefore, the domain of the function (cd)(x) will be all real values of x except x = 2. (Answer)

8 0

3 years ago

Read 2 more answers

Find a and b so that f(x) = x^3 + ax^2 + b will have a critical point at (2,3).

Digiron [165]

Using the critical point concept, it is found that a = -3 and b = 7.

<h3>What are the critical points of a function?</h3>

The critical points of a function are the values of x for which:

$f^{\prime}(x) = 0$

In this problem, the function is:

$f(x) = x^3 + ax^2 + b$

Hence, the derivative is:

$f^{\prime}(x) = 3x^2 + 2ax$

Then:

$f^{\prime}(x) = 0$

$3x^2 + 2ax = 0$

$x(3x + 2a) = 0$

$x = 0$

$3x + 2a = 0$

$3x = -2a$

$x = -\frac{2a}{3}$

Since the critical point is at x = 2, we have that:

$-\frac{2a}{3} = 2$

$-2a = 6$

$2a = -6$

$a = -3$

Then:

$f(x) = x^3 - 3x^2 + b$

Critical point at (2,3) means that when $x = 2, y = 3$ , then:

$3 = 2^3 - 3(2)^2 + b$

$8 - 12 + b = 3$

$b = 7$

You can learn more about the critical point concept at brainly.com/question/2256078

4 0

2 years ago

Cos^2x+cos^2x*cot^2x

prisoha [69]

Answer:

Cos^2x csc^2x

Step-by-step explanation:

Cos^2x (1 + cot^2x)

3 0

3 years ago

The product of two consecutive even integers is 288 find the integers.

algol [13]

Using algebra
n and n+2 are the integers
n*(n+2)=288
n^2+2n=288
n^2+2n-288=0
288=2*2*2*2*2*3*3=(2*2*2*2)*(2*3*3)=16*18 and 18-16=2
factor(n+18)(n-16)=0
n+18=0
n=-18
n+2=-16
this is one solution
n-16=0
n=16
n+2=18
this is another solution
as you can see, it's just a matter of factoring 288using arithmetic...√288=16.97≈1716 and 18 are the integers (as well as -16 and -18)

5 0

3 years ago