Find the axis of symmetry, vertex and y-intercept of the function
f(x) = -2x² + 16x + 40

The correct answer is: 40

This quadratic function is in standard form, f(x)=ax²+bx+c.
For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.
In f(x)= -2x² + 16x + 40,  a= -2, b= 16, and c= 40. So, the equation for the axis of symmetry is given by

x = −(16)/2(-2)

x = -16/-4

x = 4
The equation of the axis of symmetry for f(x)= -2x² + 16x + 40 is x = 4.
The x coordinate of the vertex is the same:

h = 4
The y coordinate of the vertex is :

k = f(h)

k = -2h² + 16h + 40

k = -2(4)² + 16(4) + 40

k = -32 + 64 + 40

k = 72
Therefore, the vertex is (4 , 72)
For finding the y- intercept we firstly rewrite the equation by substituting 0 for x.

y = -2(0)² + 16(0) + 40

y = 0 + 0 + 40

y = 40
Therefore, Axis of symmetry is x = 4
Vertex is (4 , 72)
Y- intercept is 40.