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:

Vertex = (0,-1)

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

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

Step-by-step explanation:

The given function is

$f(x)=4x^2-1$           .... (1)

The vertex form of a parabola is

$g(x)=a(x-h)^2+k$         ..... (2)

where, a is a constant, (h,k) is vertex of the parabola.

From (1) and (2) we get

$a=4,h=0,k=-1$

So, the vertex of the parabola is (0,-1).

Substitute f(x)=0 in equation (1) to find x-intercepts.

$0=4x^2-1$

Add 1 on both sides.

$1=4x^2$

Divide both sides by 4.

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

Taking square root both sides.

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

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

The x-intercepts are $-\frac{1}{2}$ and $\frac{1}{2}$.

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

Substitute x=0 in equation (1) to find the y-intercept.

$f(0)=4(0)^2-1=-1$

Therefore the y value of the y-intercept is y = -1.