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

Answer:

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

The x value of its largest x-intercept is $\frac{1}{2}$.

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

Step-by-step explanation:

A quadratic function in the format:

$f(x) = ax^{2} + bx + c$

Has the vertex $(x_{v}, y_{v})$ given by:

$x_{v} = -\frac{b}{2a}$

$y_{v} = f(x_{v})$

So

$f(x) = 4x^{2} - 1$, we have that:

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

So

$x_{v} = -\frac{b}{2a} = -\frac{0}{8} = 0$

$y_{v} = f(0) = 4*(0)^{2} - 1 = -1$

Its vertex is $(0,-1)$

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

So

$f(x) = 0$

$4x^{2} - 1 = 0$

$4x^{2} = 1$

$x^{2} = \frac{1}{4}$

$x = \pm \sqrt{\frac{1}{4}}$

$x = \pm \frac{1}{2}$

The x value of its largest x-intercept is $x = \frac{1}{2}$.

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 $y = -1$.