Solve a to e A quadratic function of the forx fx ax2 bx c

Solve a to e:

A quadratic function of the forx f(x) = ax^2 + bx + c with b^2 - 4ac greaterthan 0 may also be written in the form f(x) a(x-r_1)(x-r_2), where r_1 and r_2 are the x-intercepts of the graph of the quadratic function. Find a quadratic function whose x-intercepts are -3 and 1 with a=1, a=2, a=-2, and a = 3. How does the value of a affect the intercepts? How does the value of a affect the axis of symmetry? How does the value of a affect the vertex? Compare the x-coordinate of the vertex with the midpoint of the x-intercepts. What might you conclude?

Solution

a) f(x) = a(x-r1)(x -r2)

r1 = -3 ; r2 = 1

a= 1 ; f(x) = (x+3)(x-1) = x^2 +2x -3

a = 2 ; f(x) = 2(x+3)(x-1) = 2x^2 +4x -6

a= -2 ; f(x) = -2(x+3)(x-1) = -2x^2 -4x +6

a = 3 ; f(x) = 3(x+3)(x-1) = 3x^2 +6x -9

b) The value of a does not afect the x intercepts.However it changes the value of y intercept

c) the axis of symmetry is a vertical line x=b/2a

Since , the ratio of b/2a remains unchanged as we can see in the cases = -2/2 = -4/4 = -6/6 =-1

So, the axis of symmetry does not change with value of a

d) The vertex ( h, k) =( -b/2a , f(-b/2a) )

Th x coordinate of vertex does not chnage .However k changes with chnage in a

a = 1 ; k = -4

a = 2 ;k = -8

So, Vertex changes with chnage in value of a