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Alecsey [184]
3 years ago
9

Which function has a horizontal asymptote of y = 3?

Mathematics
2 answers:
scZoUnD [109]3 years ago
7 0
The answer is D f(x)=2(4^x)+3
Alex_Xolod [135]3 years ago
4 0

All of the four functions are exponential functions of the form:f(x)=Pa^{x}+Q

Where,

  • P changes the parent function's, f(x)=a^{x}, slope (more steep or less steep)
  • Q shifts the parent function, f(x)=a^{x}, upwards if Q is positive and downwards if Q is negative.

The parent function looks like the first image attached. Note that the x-axis (y=0) is the horizontal asymptote.

Since question asked for a function with horizontal asymptote of y=3, we can shift the function 3 units above by changing Q to +3. The fourth option has an equation of an exponential that has Q value of +3. This is the correct choice.


ANSWER: f(x)=2(4^{x})+3

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What is the solution to sqrt 17-x=x+3? Show your work.
snow_lady [41]

Answer:

x=1

Step-by-step explanation:

Remember:

(\sqrt[n]{a})^n=a\\\\(a+b)=a^2+2ab+b^2

Given the equation \sqrt{17-x}=x+3, you need to solve for the variable "x" to find its value.

You need to square both sides of the equation:

(\sqrt{17-x})^2=(x+3)^2

17-x=(x+3)^2

Simplifying, you get:

17-x=x^2+2(x)(3)+3^2\\\\17-x=x^2+6x+9\\\\x^2+6x+9+x-17=0\\\\x^2+7x-8=0

Factor the quadratic equation. Find two numbers whose sum be 7 and whose product be -8. These are: -1 and 8:

(x-1)(x+8)=0

Then:

x_1=1\\x_2=-8

Let's check if the first solution is correct:

\sqrt{17-(1)}=(1)+3

4=4 (It checks)

Let's check if the second solution is correct:

\sqrt{17-(-8)}=(-8)+3

5\neq-5 (It does not checks)

Therefore, the solution is:

x=1

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