Answer:
Plot these points
(-3,0)
(-5,0)
(-4,1)
(-2,-3)
(-6,-3)
Step-by-step explanation:
Let find the zeros of the equation.
We can factor this equation.
Factor out the -1.
[tex] - 1( {x}^{2} + 8x + 15)[/tex]
Factor using AC method
[tex] - 1(x + 3)(x + 5)[/tex]
Set all the terms equal to zero.
[tex]x + 3 = 0[/tex]
[tex]x + 5 = 0[/tex]
[tex]x = - 3[/tex]
[tex]x = - 5[/tex]
So our x intercepts are -3,0 and -5,0.
To find our vertex, apply the -b/2a.
[tex] \frac{8}{ - 2} = - 4[/tex]
Then
Substitute-4 for x.
[tex]y = - {4}^{2} - 8( - 4) - 15 = 1[/tex]
So our vertex is at (-4,1).
Find some other points like -2 and -6.
To find -2, substitute-2 into the quadratic.
[tex]y = - ( { - 2}^{2}) - 8( - 2) - 15 = - 3[/tex]
So -2,-3.
Since y=-4 is the axis of symmetry
-6,-3
