Answer:
[tex]f(x) = -x^2 - 2x + 8[/tex]
y-intercept is when x = 0:
[tex]\implies f(0) = -(0)^2 - 2(0) + 8=8[/tex]
So the y-intercept is (0, 8)
x-intercepts are when f(x) = 0:
[tex]\implies f(x)=0[/tex]
[tex]\implies -x^2 - 2x + 8=0[/tex]
[tex]\implies x^2 + 2x - 8=0[/tex]
[tex]\implies x^2 - 2x + 4x - 8=0[/tex]
[tex]\implies x(x - 2) + 4(x-2)=0[/tex]
[tex]\implies (x+4)(x - 2)=0[/tex]
Therefore:
[tex]\implies (x+4)=0 \implies x=-4[/tex]
[tex]\implies (x-2)=0 \implies x=2[/tex]
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 [tex]\sf \dfrac{x_1+x_2}{2}=\dfrac{-4+2}{2}=-1[/tex]
Substituting x = -1 into the function:
[tex]\implies f(-1) = -(-1)^2 - 2(-1) + 8=9[/tex]
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 [tex]f(x)\leq 9[/tex]
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:
[tex](x, -x^2 - 2x + 8)[/tex] for all real numbers of x
