Question

Consider the Quadratic function f(x) = 4x^2 - 1. Its vertex is Preview The x value of its largest x-intercept is x = The y value of the y-intercept is y = Preview Preview Preview

Answer 1

Its vertex is
`(0,-1)`.

The x value of its largest x-intercept is
`(1)/(2)`.

The y value of the y-intercept is
`y = -1`.

Explanation:

A quadratic function in the format:

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

Has the vertex
`(x_(v), y_(v))` given by:

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

`y_(v) = f(x_(v))`

So

`f(x) = 4x^(2) - 1`, we have that:

`a = 4, b = 0, c = -1`

So

`x_(v) = -(b)/(2a) = -(0)/(8) = 0`

`y_(v) = f(0) = 4*(0)^(2) - 1 = -1`

Its vertex is
`(0,-1)`

The x-intercept values are the values of x when f(x) = 0;

So

`f(x) = 0`

`4x^(2) - 1 = 0`

`4x^(2) = 1`

`x^(2) = (1)/(4)`

`x = \pm \sqrt{(1)/(4)}`

`x = \pm (1)/(2)`

The x value of its largest x-intercept is
`x = (1)/(2)`.

The y-intercept is the value of f(x) when x = 0, so, f(0)

`f(x) = 4x^(2) - 1`

`f(0) = 4*(0)^(2) - 1 = 0-1 = -1`

The y value of the y-intercept is
`y = -1`.