Answer:
Following are the solution to the gvien question:
Step-by-step explanation:
Let the quadratic equation is:
[tex]\to h(x) = (x + 1)^2 - 4[/tex]
vertex is:
[tex]\to h(x) = a(x -h)^2 + k[/tex]
(h) = axis of symmetry
(h,k) = vertex.
By using the given equation:
[tex]h(x) = (x - (-1))^2 - 4[/tex]
Hence,
[tex]h = -1 \\\\ k = -4[/tex]
line of symmetry [tex]x = -1[/tex]
vertex is [tex](h,k) = (-1,-4)[/tex]
finding the x intercept:
[tex](x + 1)^2 = 4\\\\\sqrt{(x + 1)^2} = \sqrt{4}\\\\x + 1 = \pm 2\\\\x = 2-1 \ \ \ \ \ \ \ \ \ \ or \ \ \ \ \ \ \ \ \ \ \ \ \ \ x =-2 -1\\\\x = 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ or \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x =-3\\\\[/tex]
x -intercepts -3,1
Calculating the y-intercept when x = 0 putting into the real equation:
[tex]h(x) = (0 +1)^2 - 4 \\\\y = 1 - 4\\\\y = -3[/tex]
Please find the graph file in the attachment.
