Result: 4020342569677260278
Answer:
(f-g)(x)=-x^(2)+2x+8
the solutions are:
<em><u>x=4 or x=-2</u></em>
Step-by-step explanation:
(f-g)(x)=2x+1-(x^(2)-7)
(f-g)(x)=-x^(2)+2x+1+7
(f-g)(x)=-x^(2)+2x+8
does this help or should I solve for the zeros/solutions of this quadratic equation?
then:
-x^(2)+2x+8=0
-(x^(2)-2x-8)=0
x^(2)-2x-8=0
(x-4)(x+2)=0
<em><u>x=4 or x=-2</u></em>
Equation of the parabola: y = ax^2 + bx + c. Find a, b, and c.
x of axis of symmetry:
x
=
−
b
2
a
=
3
-> b = -6a
Writing that the graph passing at point (1, 0) and point (4, -3):
(1) 0 = a + b + c -> c = - a - b = - a + 6a = 5a
(2) -3 = 16a + 4b + c --> -3 = 16a - 24a + 5a = -3a --> a = 1
b = -6a = -6; and c = 5a = 5
y
=
x
2
−
6
x
+
5
Check with x = 1: -> y = 1 - 6 + 5 = 0. OK
Cos 45 = √6 / y
√2 / 2 = √6 / y
Inverse both sides
2 / √2 = y / √6
Multiply both sides by √6
√6 × 2 / √2 = √6 × y / √6
√2 × √3 × 2 / √2 × 1 = y
<h2>y = 2√3</h2>
