## Precalculus (6th Edition)

$5$; two
RECALL: The quadratic function $f(x) = ax^2+bx+c$ opens: (a) downward when $a \lt0$; or (b) upward when $a\gt0$. The given quadratic function has $a=-2$. Thus, its graph is a parabola that opens downward. The y-intercept of the function can be found by setting $x=0$ then solving for $f(0)$: $f(x)=-2x^2-6x+5 \\f(0) = -2(0^2)-6(0)+5 \\f(0) = 5$ The y-intercept is $(0, 5)$. The parabola opens downward. Having a y-intercept of $(0, 5)$, which is above the x-axis, means that the vertex is also above the x-axis. A parabola that opens downward and whose vertex is above the x-axis will cross the x-axis at two points. Thus, the function has 2 x-intercepts. Therefore, the missing expressions in the given statement are (in order): $5$; two