Question

Assist me with this graph.

Answer 1

3.1.1. A(-4, 0) B(0, -2) E(4, 0)

3.1.2. k = -16

3.1.3. (-∞, ∞)

3.1.4. [-16, ∞)

3.1.5. [-6, 4]

Explanation:

Given functions:

`\begin{cases}f(x)=x^2+k\\g(x)=-2x+8\\h(x)=(6)/(x-2)+1\end{cases}`

Question 3.1.1.

Points A and B are the x-intercept and y-intercept of function h(x).

Point E is the x-intercept of function g(x).

The x-intercept of function h(x) is when h(x) = 0:

`\begin{aligned}h(x) & = 0\\\implies (6)/(x-2)+1 & = 0\\(6)/(x-2) & = -1\\6 & = -1(x-2)\\6 & = -x+2\\4 & = -x\\x & = -4\end{aligned}`

Therefore, the coordinates of point A are (-4, 0).

The y-intercept of function h(x) is when x = 0:

`\begin{aligned}h(x) & = (6)/(x-2)+1\\\implies h(0) & = (6)/(0-2)+1\\ & = -3+1\\& = -2\end{aligned}`

Therefore, the coordinates of point B are (0, -2).

The x-intercept of function g(x) is when g(x) = 0:

`\begin{aligned}g(x) & = 0\\\implies -2x+8 & = 0\\-2x & = -8\\x & = 4\end{aligned}`

Therefore, the coordinates of point E are (4, 0).

Question 3.1.2.

As function f(x) passes through point C (-6, 20), substitute the point into the equation of the function to find the value of k:

`\begin{aligned}f(-6) & = 20\\\implies (-6)^2 + k & = 20\\36 + k & = 20\\k & = 20 - 36\\k & = -16\end{aligned}`

Hence, proving that the value of k is -16.

Question 3.1.3.

The domain is the set of all possible input values (x-values).

The domain of function f(x) is unrestricted, therefore its domain is:

• Solution: -∞ < x < ∞
• Interval notation: (-∞, ∞)

Question 3.1.4.

The range is the set of all possible output values (y-values).

The range of function f(x) is restricted, since the minimum point of the parabola is at (0, -16). Therefore its range is:

• Solution: f(x) ≥ -16
• Interval notation: [-16, ∞)

Question 3.1.5.

`\begin{aligned}g(x) - f(x) & \geq 0\\\implies g(x) &\geq 0+f(x)\\ \implies g(x) & \geq f(x)\end{aligned}`

Therefore, find the interval for which g(x) is greater than or equal to f(x).

From inspection of the given graph this is between points C and E:

• Solution: -6 ≤ x ≤ 4
• Interval notation: [-6, 4]