Answer:
[tex]y=x^2+8x+15[/tex]
Intercept form is y= (x+5)(x+3)
vertex is (-4,1)
y intercept is (0,15)
x intercepts are (-5,0) (-3,0)
Step-by-step explanation:
Given quadratic function
[tex]y=(x+4)^2-1[/tex]
To rewrite it in standard form we need to expand (x+4)^2
[tex]y=(x+4)(x+4)-1[/tex]
[tex]y=x^2+8x+16-1[/tex]
[tex]y=x^2+8x+15[/tex]
Intercept form is y=a(x-p)(x-q)
Factor x^2 + 8x + 15
product is 15 and sum is 8, 5*3= 15 and 5+3=8
(x+5)(x+3)
Intercept form is y= (x+5)(x+3)
Vertex form of equation is y=(x-h)^2+k
where (h,k) is the vertex
[tex]y=(x+4)^2-1[/tex], here h = -4 and k =1 so vertex is (-4,1)
To find y intercept plug in 0 for x
y= (x+5)(x+3)
y= (0+5)(0+3) = 15, so y intercept is (0,15)
to find x intercept plug in 0 for y
y= (x+5)(x+3)
0= (x+5)(x+3), x+5 =0 and x+3=0
x=-5 and x= -3
So x intercepts are (-5,0) (-3,0)
