Question

How do you find the y intercept, axis of symmetry and the vertex to graph the function `f(x)=1/2x^2+3x+9/2`?

Answer 1

`y`-intercept is `9/2`, axis of symmetry is vertical along `x=-3`, and vertex is `(-3,0)`.

Explanation:

A polynomial of degree 2 such as this represents a parabola. Given that the squared term is `x^2`, we know that this is a vertical parabola. In Precalculus, to find the items asked, you need to rearrange the function into the standard parabola form of:

`(x-x_0)^2 = 4a(y-y_0)`

(With vertex at `(x_0,y_0)` and focus distance of `a`).

To do this, we have to rewrite `f(x)` as `y`, then complete the square on the `x` terms, and rearrange into the standard form:

`y = 1/2 x^2 + 3x + 9/2`

`2y = x^2 + 6x + 9`

`2y = (x+3)^2`

`(x+3)^2 = 4*1/2*(y-0)`

From this form, we can directly read that the vertex is at `(-3,0)` and the focus distance is `1/2`. Thus:

• The `y`-intercept happens when `x = 0`. Plug in `x=0` and you get `9/2`.
• Since this is a vertical parabola, the axis-of-symmetry is a vertical line which passes through the vertex and focus. It will lie along the vertical line `x=-3`
• The vertex is given from the standard form as `(-3,0)`

graph{1/2x^2+3x+9/2 [-12.87, 7.13, -1.64, 8.36]}

Calculus Side Note

Once you learn Calculus, you can dispense with putting the parabola into the standard form in order to find the vertex. The vertex by definition will be either the minimum point of the parabola (for `ax^2` forms) or the maximum point of the parabola (for `-ax^2` forms).

Thus, a quick derivative set equal to 0 will yield the x point for the vertex because it will identify the single critical point:

`f'(x) = 2*1/2*x + 3 = x + 3`

`f'(x) = 0 -> x+3 = 0 :. x = -3`

Evaluate `f(x)` at this point to get the `y`-coordinate for the vertex. (Note: The result of the derivative work is also technically the equation for the axis of symmetry line.)

Answer 2

`y=9/2` --- `y` intercept
Vertex `(-3,0)`
Axis of symmetry `x=-3`

Explanation:

Given -

`f(x)=1/2 x^2+3x+9/2`

Find `y` intercept. Put `x=0`

`y=1/2 x^2+3x+9/2`
`y=1/2(0^2)+3(0)+9/2=9/2`

`y=9/2`

Vertex

`x=(-b)/(2a)=(-3)/(2xx1/2)=-3`

At `x=-3`

`y=1/2(-3)^2+3(-3)+9/2`
`y=(-9)/2-9+9/2`
`y=(9-18+9)/2=(0)/2=0`
Vertex `(-3,0)`

Axis of symmetry

`x=-3`