Question

y = 2x^2 + 8x - 5 find max or min

Answer 1

We can convert this to vertex form to find the vertex (max or min):
2x^2 + 8x -5
2(x^2+8x)-5

Use the completing the square method:
(8/2)^2 gives you your c value

2(x^2 + 8x + 16 - 16)-5
move the -16 out of the brackets
2(x^2 + 8x +16)-37
factor the trinomial

2(x+4)^2 -37

Answer 2

so how do I find the min or the max plz

Answer 3

Oops sorry! So in vertex form
a(x-p)^2 +q
p is the x coordinate of the vertex and q is the y coordinate

So if we have 2(x+4)^2 -37
we really have 2(x--4)^2 -37

So our x coordinate is -4 and our y coordinate is -37.

Our vertex is at (-4, -37), and because the a value is positive, the parabola opens upwards and has a minimum value.