Answer
The equation when graphed, is a parabola with vertex at $(2,14)$.
Work Step by Step
$$y = -x^{2} + 4x + 10$$
Since the coefficient of the first term is negative, then the parabola opens downward. The next thing we do is to determine whether the equation has real solution, and that intercepts can be computed. We can do this by computing for the discriminant using the formula: $$b^{2}−4ac$$, with a=-1 b=4 and c=10
Thus, $$b^{2}−4ac$$ $$= 4^{2}−4(-1)(10) $$=56$$
Since the discriminant is positive, we can now proceed with computing for the intercepts. If we set $x=0$, then $y=10$
We now use the quadratic formula to compute for the x-intercepts using a=-1 b=4 and c=10. $$x=-b±\frac{\sqrt b^{2}-4ac}{2a}$$ $$x=-4±\frac{\sqrt 4^{2}-4(-1)(10)}{2(-1)} $$x=-1.741657387$$ $$x=5.741657387$$
We now have three points to graph: $(-1.741657387,0)$, $(0,10)$, $(5.741657387,0)$.
Now we compute for the vertex using the formula: $$x=\frac{-b}{2a}$$ $$x=\frac{-4}{2(-1)}$$ $$x=2$$
Substituting this to the original equation: $$y = -x^{2} + 4x + 10$$ $$y=-(2)^{2} +4(2) + 10$$ $$y = 14$$
Hence, the vertex is at $(2,14)$. When graphed, a parabola which indeed opens downward will be formed.
