Answer:
The horizontal distance is 5
Explanation:
The zeros of a function f(x) are the values of x which:
[tex]f(x)=0[/tex]
In this case, we have the function:
[tex]-x^2+x+6[/tex]
We want to find the zeros:
[tex]-x^2+x+6=0[/tex]
Now, we can use the quadratic formula:
[tex]x_{1,2}=\frac{-1\pm\sqrt{1^2-4(-1)6}}{2(-1)}[/tex][tex]x_{1,2}=\frac{-1\pm\sqrt{1+4\cdot6}}{-2}[/tex][tex]x_{1,2}=\frac{1\pm\sqrt{1+24}}{2}[/tex][tex]x_{1,2}=\frac{1\pm\sqrt{25}}{2}[/tex][tex]x_{1,2}=\frac{1\pm5}{2}[/tex]
Then:
[tex]\begin{gathered} x_1=\frac{1+5}{2}=\frac{6}{2}=3 \\ \end{gathered}[/tex][tex]x_2=\frac{1-5}{2}=\frac{-4}{2}=-2[/tex]
The two roots are x = -2 and x = 3
To find the distance, we take the absolute value of the difference:
[tex]Distance=|-2-3|=|-5|=5[/tex]
The distance is 5
