You use permutation formula
n!/(n-r)!
n = 5
r = 3
5!/(5-3)!
= 120/2=60
the equation of a parabola in
vertex form
is.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
∣
∣
∣
2
2
y
=
a
(
x
−
h
)
2
+
k
2
2
∣
∣
∣
−−−−−−−−−−−−−−−−−−−−−
where
(
h
,
k
)
are the coordinates of the vertex and a
is a multiplier
to obtain this form
complete the square
y
=
x
2
+
2
(
4
)
x
+
16
−
16
+
14
⇒
y
=
(
x
+
4
)
2
−
2
←
in vertex form
⇒
vertex
=
(
−
4
,
−
2
)
to obtain the intercepts
∙
let x = 0, in the equation for y-intercept
∙
let y = 0, in the equation for x-intercept
x
=
0
⇒
y
=
0
+
0
+
14
=
14
←
y-intercept
y
=
0
⇒
(
x
+
4
)
2
−
2
=
0
←
add 2 to both sides
⇒
(
x
+
4
)
2
=
2
take the square root of both sides
√
(
x
+
4
)
2
=
±
√
2
←
note plus or minus
⇒
x
+
4
=
±
√
2
←
subtract 4 from both sides
⇒
x
=
−
4
±
√
2
←
exact values
graph{(y-x^2-8x-14)((x+4)^2+(y+2)^2-0.04)=0 [-10, 10, -5, 5]}
We have to find horizontal component of Mayosons velocity.
#a
#b
Here 2 vectors A=8 and B=2.5
Please, use " ^ " to denote exponentiation: p(t) = t^2 + 5t + 6.
To find the critical points, differentiate p(t) with respect to t, set the result = to 0, and then solve the resulting equation for t:
p '(t) = 2t + 5 = 0
Solving for t: 2t = -5, and so t = -5/2. (-5/2, p(-5/2)) is the critical point. That evaluates to (-5/2, -0.25). This happens to be the vertex of a parabola that opens up.
