Answer
$5$;
two
Work Step by Step
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
