Answer
x-intercepts: $-2$ and $2$
y-intercept: none
Work Step by Step
RECALL:
(1) A point where the graph touches or crosses the x-axis is called an x-intercept. Given an equation, the x-intercept/s can be found by setting $y=0$ then solving for $x$.
(2) A point where the graph touches or crosses the y-axis is called a y-intercept.
Given an equation, the y-intercept/s can be found by setting $x=0$ then solving for $y$.
Solve for the x-intercept/s by setting $y=0$ then solving for $x$:
$\begin{array}{ccc}
&25x^2-y^2 &= &100
\\&25x^2-0^2) &= &100
\\&25x^2 &= &100
\\&\dfrac{25x^2}{25} &= &\dfrac{100}{25}
\\&x^2 &= &4
\\&x &= &\pm\sqrt{4}
\\&x &= &\pm\sqrt{2^2}
\\&x &= &\pm 2
\end{array}$
The x-intercepts are $-2$ and $2$.
Solve for the y-intercept/s by setting $x=0$ then solving for $y$:
$\begin{array}{ccc}
&25x^2-y^2 &= &100
\\&25(0^2)-y^2 &= &100
\\&-y^2 &= &100
\\&-1(-y^2) &= &-1(100)
\\&y^2 &= &-100
\end{array}$
Note that there is no real number whose square negative.
Thus, there is no real solution to the equation. above.
This means that the graph of the function has no y-intercept.
