Question

For the quadratic function `f(x)= 2x^2 - 11`, how do you find the y intercept, the equation of the axis of symmetry and the x-coordinate of the vertex?

Answer 1

Refer to the explanation.

Explanation:

Given:

`f(x)=2x^2-11` is in standard form:

`f(x)=ax^2+bx+c`,

where:

`a=2`, `b=0`, `c=-11`.

Substitute `y` for `f(x)`.

`y=2x^2-11`

The y-intercept is the value of `y` when `x=0`, which basically means that the value for `c` is the y-intercept.

`y=2(0)^2-11`

`y=-11`

The axis of symmetry is the vertical line that divides the parabola into two equal halves. It is also the x-coordinate of the vertex. The equation is:

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

Plug in the known values.

`x=(-0)/(2*2)`

`x=0`

The vertex is the minimum or maximum point of a parabola. Substitute `0` for `x` and solve for `y`.

`y=2(0)^2-11`

`y=-11`

The vertex is `(0,-11)`

graph{y=2x^2-11 [-15.35, 16.68, -15.64, 0.38]}