Provide an appropriate response.Given the solutions of a quadratic equation, is it possible to reconstruct the original equation? Why or why not?
What will be an ideal response?
No. Given the solutions of a quadratic equation, it is possible to find an equation equivalent to the original equation but not necessarily expressed in the same form as the original equation. For example, we can find a quadratic equation with solutions -2 and 4.
(x - (-2)) (x - 4) = 0
(x + 2)(x - 4) = 0
x2 - 2x - 8 = 0
Now x2 - 2x - 8 = 0 has solutions -2 and 4. However, the original equation might have been in another form, such as 
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Provide an appropriate response.Write an equation of the tangent line to the graph of y = 
at the point where x = 0.
A. y = - 6x + 4 B. y = - 6x - 4 C. y = 6x - 4 D. y = 6x + 4
Write the product as a sum or difference containing only sines or cosines.sin(5?) sin(9?)
A. sin2(45?2)
B. 
[cos(4?) - cos(14?)]
C. 
[cos(14?) - sin(4?)]
D. 
[- cos(4?) - cos(14?)]
Find the vertex form for the quadratic function. Then find each of the following:(A) Intercepts(B) Vertex(C) Maximum or minimum(D) Rangef(x) = x2 + 4x + 3
A. Standard form: f(x) = (x - 2)2 - 1 (A) x-intercepts: 1, 3; y-intercept: 3 (B) Vertex (-2, -1) (C) Minimum: -1 (D) y ? -1 B. Standard form: f(x) = (x + 2)2 - 1 (A) x-intercepts: - 3, -1; y-intercept: 3 (B) Vertex (-2, -1) (C) Minimum: -1 (D) y ? -1 C. Standard form: f(x) = (x - 2)2 - 1 (A) x-intercepts: - 3, -1; y-intercept: 3 (B) Vertex (-2, -1) (C) Maximum: -1 (D) y ? -1 D. Standard form: f(x) = (x + 2)2 - 1 (A) x-intercepts: - 3, -1; y-intercept: 3 (B) Vertex (2, -1) (C) Minimum: -1 (D) y ? -1
Perform the indicated operation. Write your result in a + bi form.-5i(-5i)
A. -25i B. -25i2 C. -25 D. 25