Question

Find the vertex, axis of symmetry, and y-intercept of the quadratic function.

Answer 1

see explanation

Explanation:

given a quadratic function in standard form

f(x) = ax² + bx + c ( a ≠ 0 ) , then

the x- coordinate of the vertex is

`x_(vertex)` = -
`(b)/(2a)`

f(x) =
`(1)/(2)` x² + 4x ← is in standard form

with a =
`(1)/(2)` , b = 4 , c = 0 , then

`x_(vertex)` = -
`(4)/(1)` = - 4

substitute x = - 4 into f(x) for corresponding y- coordinate of vertex

f(- 4) =
`(1)/(2)` (- 4)² + 4(- 4) =
`(1)/(2)` × 16 - 16 = 8 - 16 = - 8

vertex = (- 4, - 8 )

the equation of the axis of symmetry is a vertical line passing through the vertex with equation

x = c ( c is the x- coordinate of the vertex ) , then equation of axis of symmetry is

x = - 4

to find the y- intercept substitute x = 0 into f(x)

f(0) =
`(1)/(2)` (0)² + 4(0) = 0 + 0 = 0

then y- intercept is 0

Answer 2

Below in bold.

Explanation:

f(x) = 1/2x^2 + 4x

Convert to vertex form:

f(x) = 1/2(x^2 + 8x)

= 1/2 [(x + 4)^2 - 16]

= 1/2(x + 4)^2 - 8 <------ Vertex form.

So the Vertex is at (-4, - 8).

Axis of symmetry is x = -4.

y -intercept (when x = 0) = 8-8 = 0.