Question

The graph of f(x) = -x2 - 2x + 8 is shown. Which of the following describes all solutions for f(x)?

Answer 1

Answer:

(x, x2 -2x+8) where −5 ≤ x ≤ 3

Step-by-step explanation:

The given function is  f(x)=x2- 2x+ 8

We want to select the option that describes all the solutions to the parabola.

The domain of the parabola is −5 ≤ x ≤ 3.

This means that  any x=a on −5 ≤ x ≤ 3 that satisfies (a,f(a)), is a solution.

This can be rewritten as (a,-a2- 2a+ 8)

Therefore for x belonging to −5 ≤ x ≤ 3, all solutions are given by:

(x, x2 -2x+8) where −5 ≤ x ≤ 3.

Answer 2

Answer:

$f(x) = -x^2 - 2x + 8$

y-intercept is when x = 0:

$\implies f(0) = -(0)^2 - 2(0) + 8=8$

So the y-intercept is (0, 8)

x-intercepts are when f(x) = 0:

$\implies f(x)=0$

$\implies -x^2 - 2x + 8=0$

$\implies x^2 + 2x - 8=0$

$\implies x^2 - 2x + 4x - 8=0$

$\implies x(x - 2) + 4(x-2)=0$

$\implies (x+4)(x - 2)=0$

Therefore:

$\implies (x+4)=0 \implies x=-4$

$\implies (x-2)=0 \implies x=2$

So the x-intercepts are (-4, 0) and (2, 0)

The vertex is the turning point of the parabola.  The x-value of the vertex is the x-value between the 2 zeros (x-intercepts).

Therefore, the x-value of the vertex is $\sf \dfrac{x_1+x_2}{2}=\dfrac{-4+2}{2}=-1$

Substituting x = -1 into the function:

$\implies f(-1) = -(-1)^2 - 2(-1) + 8=9$

So the vertex is (-1, 9)

The domain is the input values, so x = all real numbers.

The range is the output values, so the range is $f(x)\leq 9$

Therefore, (-4, 0), (-1, 9), (0, 8) and (2, 0) are solutions of the function, HOWEVER they are not ALL the solutions.

All solutions of the function are:

$(x, -x^2 - 2x + 8)$ for all real numbers of x