## College Algebra 7th Edition

x-intercepts: $-2$ and $2$ y-intercept: none
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.