Answer:
[tex] 0 = (x+1)(x-5)[/tex]
[tex] x= -1 , x= 5[/tex]
[tex] f(x) = x^2 -4x -5[/tex]
[tex] V_x = -\frac{b}{2a}[/tex]
With [tex] a = 1, b=-4 , c=-5[/tex]
[tex] V_x = -\frac{-4}{2*1}= 2[/tex]
[tex]f(2) = 2^2 -4*2 -5 = -9[/tex]
[tex] f(0) = 0^2 -4*0 -5=-5[/tex]
So then the x intercept would be (0,-5). And finally we can graph the function as we can see in the figure attached.
Step-by-step explanation:
For this case we know the following function:
[tex] f(x) = (x+1)(x-5)[/tex]
We can begin the zeros or the values where the function is 0 like this:
[tex] 0 = (x+1)(x-5)[/tex]
And solving for x we got:
[tex] x= -1 , x= 5[/tex]
Now we can rewrite the expression like this:
[tex] f(x) = x^2 -4x -5[/tex]
And we can find the position for the vertex at x with this formula:
[tex] V_x = -\frac{b}{2a}[/tex]
With [tex] a = 1, b=-4 , c=-5[/tex]
And replacing we got:
[tex] V_x = -\frac{-4}{2*1}= 2[/tex]
And then with the coordinate of x for the vertex we can find the coordinate of y replacing the value of x obtained for the vertex.
[tex]f(2) = 2^2 -4*2 -5 = -9[/tex]
Then we can find the intercept using the value of x=0 and replacing into the function we got:
[tex] f(0) = 0^2 -4*0 -5=-5[/tex]
So then the x intercept would be (0,-5). And finally we can graph the function as we can see in the figure attached.
