What is the vertex of the function f(x)=4x^2+16x+14?

Answer 1

y = 4x^2 + 16x + 14

= 2(2x^2 + 8x + 7)

=4(x^2 + 4x + 7/2)

= 4(x^2 + 4x + 4) -2

= 4(x+2)^2 -2

=a(x-h)^2 + k where (h,k) = vertex = (-2, -2)

vertex is (-2, -2)

or take the derivative and set equal to zero

y' = 8x + 16 = 0

x =-16/8 = -2

y = 4(-2)^2 + 16(-2) + 14 = -2

Answer 2

f(x) = 4(x² + 4x + 4) - 16 + 14 = 4(x + 2)² - 2.

So, vertex at point (- 2, - 2)

Answer 3

You can find the x coordinate of the vertex using the formula -b/(2a)

x vertex = -16/(2*4) = -2

To find the y coordinate, plug in the x value into the original equation

f(x) = 4x²+16x+14

f(-2) = 4(-2)²+16(-2)+14

f(-2) = 16-32+14 = -2

The vertex would be (-2,-2)