Sketch the graph of the function y=-2x^2-4x-6

Question

Please also state the vertex and the line of symmetry .Vertex

                                                                              line of symmertry

Please show steps thank you so much

Answer 1

y = -2x^2 - 4x - 6 <== Standard Form

y-intercept = y(0) = –6; so (0,–6) is on parabola.

y = a x^2 translated by <h,k> gives us

y - k = a (x - h)^2, or

y = a (x - h)^2 + k <== Vertex Form

because (h,k) is the Vertex.

Expanding the Vertex Form:

y = a (x^2 – 2hx + h^2) + k

y = a x^2 – 2ah x + ah^2 + k, and comparing to Standard Form:
y = -2x^2 - 4x - 6, we see that:

a = –2
–2ah = –4 ==> h = –4/(-2(-2)) = –1
ah^2 + k = –6 ==> k = –6 – (–2)(–1)^2 = –4

So:
y = –2(x – (–1))^2 + (–4)
y = –2(x + 1)^2 – 4 <== Vertex Form

Vertex: (–1,–4)
Axis of Symmetry: x = h = –1

Since Vertex is below the x-axis and the parabola is opening down (a < 0), the parabola does not intersect the x-axis. No x-intercepts.

Corresponding point to y-intercept point of (0,–6):

(–2)––––(–1)––––(0)

(–2,–6) is corresponding point to (0,–6)

Using Vertex Form and 3 point you now have you should be able to sketch the parabola easily.