The problem asks us to find the $x$-intercepts and the $y$-intercept of the quadratic function $f(x) = 16x^2 - 49$. Specifically, we need to find the smallest $x$-intercept, the largest $x$-intercept, and the $y$-intercept.

1. Problem Description

The problem asks us to find the $x$-intercepts and the $y$-intercept of the quadratic function $f(x) = 16x^2 - 49$. Specifically, we need to find the smallest $x$-intercept, the largest $x$-intercept, and the $y$-intercept.

2. Solution Steps

To find the $x$-intercepts, we need to find the values of $x$ when $f(x) = 0$.

$16x^2 - 49 = 0$

$16x^2 = 49$

$x^2 = \frac{49}{16}$

$x = \pm \sqrt{\frac{49}{16}}$

$x = \pm \frac{7}{4}$

So, the $x$-intercepts are $x = \frac{7}{4}$ and $x = -\frac{7}{4}$.

The smallest $x$-intercept is $-\frac{7}{4}$.

The largest $x$-intercept is $\frac{7}{4}$.

To find the $y$-intercept, we need to find the value of $f(x)$ when $x = 0$.

$f(0) = 16(0)^2 - 49$

$f(0) = 0 - 49$

$f(0) = -49$

So, the $y$-intercept is $y = -49$.

3. Final Answer

The $x$ value of its smallest $x$-intercept is $x = -\frac{7}{4}$.

The $x$ value of its largest $x$-intercept is $x = \frac{7}{4}$.

The $y$ value of the $y$-intercept is $y = -49$.