Question

How do you graph the function, label the vertex, axis of symmetry, and x-intercepts. `x-4 = 1/4 (y+1)^2`?

Answer 1

The vertex is at (`4,-1`).
The axis of symmetry is `y = -1`.
The `x`-intercept is (`17/4,0`).
There are no `y`-intercepts.

Explanation:

The standard form for the equation of a parabola is

`y= a^2 + bx +c`

Your equation is

`x−4 = 1/4(y+1)^2`

We must get this into standard form.

`x−4=1/4(y^2 + 2y + 1)`

`x−4=1/4y^2 + 1/4(2y)+ 1/4`

`x−4=1/4y^2 + 1/2y + 1/4`

`x=1/4y^2 + 1/2y + 1/4 + 4`

`x=1/4y^2 + 1/2y + 17/4`

This is standard form, but with `x` and `y` interchanged.

We are going to get a sideways parabola.

`a = 1/4`, `b = 1/2`, and `c = 17/4`.

Vertex

Since `a > 0`, the parabola opens to the right.

The `y`-coordinate of the vertex is at

`y = –b/(2a) = -(1/2)/(2×(1/4)) = -(1/2)/(1/2) = -1`.

Insert this value of `x` back into the equation.

`x=1/4y^2 + 1/2y + 17/4`

`x=1/4(-1)^2 + 1/2(-1) + 17/4 = 1/4×1 -1/2 +17/4`

`x = 1/4 -1/2 + 17/4 = (1 -2 +17)/4 = 16/4 = 4`

The vertex is at (`4, -1`).

Axis of symmetry

The axis of symmetry must pass through the vertex, so

The axis of symmetry is `y = -1`.

`x`-intercept

To find the `x`-intercept, we set `y = 0` and solve for `x`.

`x=1/4y^2 + 1/2y + 17/4 = 1/4×0^2 + 1/2×0 + 17/4 = 0 + 0 + 17/4 = 17/4`

The `x`-intercept is at (`17/4,0`).

`y`-intercepts

To find the `y`-intercepts, we set `x = 0` and solve for `y`.

`x=1/4y^2 + 1/2y + 17/4`

`0=1/4y^2 + 1/2y + 17/4`

`0=y^2 + 2y + 17`

`y = (-b ± sqrt(b^2-4ac))/(2a)`

The discriminant `D = b^2 – 4ac = 4^2 – 4×1×17= 8 – 68 = -60`

Since `D < 0`, there are no real roots.

There are no `y`-intercepts.

Graph

Now we prepare a table of `x` and `y` values.

The axis of symmetry passes through `y = -1`.

Let's prepare a table with points that are 5 units on either side of the axis, that is, from `y = -6` to `y = 4`.

Plot these points.

And we have our graph. The red line is the axis of symmetry.