Question

Anyone know how to find the vertex of f(x)=2x^2+4x-16 ?

Answer 1

when you have the equation in standard form (as you do)

y = ax^2 + bx + c

the x value of the vertex is at x = -b/(2a)

the y value of the vertex you find by plugging that value of x into the equation

Answer 2

we have a = 2, b = 4, c = -16

x = -(4)/(2*2) = -4/4 = -1

what will the y value be?
y = 2(-1)^2 + 4(-1) - 16 =

Answer 3

you can also find the vertex by rearranging into vertex form, which is
y = a(x-h)^2 + k
the vertex will be at (h,k)

Answer 4

factor first. 2x^2 + 8x - 4x -16
2x (x+4) -4 (x + 4)
y= (2x-4)(x+4) your 2 roots are 2x-4= 0 2x=4 x=2 and x+4 = 0 x=-4
therefore h = 2-4 /2 h= -1
sub -1 as your x values into the equation to find y
y= (2(-1) -4) (-1 +4)
y= (-6)(3)
y=-18

Answer 5

therefore your vertex is (-1, -18)